WAEC SSCE - Physics - 1990

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Work done (W) is defined as the product of force (F) and displacement (s) in the direction of force. In the given diagram, the force F is acting at an angle ? to the direction of displacement, and the displacement is given by r. Therefore, the work done can be calculated as follows: W = F x r cos? Here, F represents the magnitude of the force, r represents the displacement of the body, and cos? represents the cosine of the angle between the force and displacement vectors. Therefore, the correct option for the work done is Fx cos?.

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Viscosity is a measure of a liquid's resistance to flow, or how "thick" or "sticky" it is. The viscosity of a liquid depends on the nature of the liquid, the temperature of the liquid, and the relative velocity between the liquid layers. However, it does not depend on the area of the surfaces in contact or the normal reaction between the liquid layers. To put it simply, imagine you have two identical cups filled with the same liquid. If you pour the liquid from one cup to the other, you will notice that the liquid flows more slowly than water. This is because the liquid has a higher viscosity. The viscosity depends on the type of liquid (e.g. honey is more viscous than water), and how fast or slow the liquid is moving (e.g. syrup flows more slowly than water). The viscosity also changes with temperature, as higher temperatures tend to decrease the viscosity of a liquid. However, the area of the surfaces in contact and the normal reaction between the liquid layers do not affect the viscosity of a liquid. These factors might affect how the liquid flows, but they don't change the liquid's intrinsic property of viscosity.

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In order to obtain a parallel beam of light from the headlamp of a car using a spherical mirror, the light source should be placed at the focal point of the mirror. This is because a spherical mirror can converge or diverge the light rays depending on where the light source is placed relative to the mirror. If the light source is placed at the focal point of the mirror, the light rays will be reflected in parallel to each other, creating a parallel beam of light. Placing the light source at any other position, such as the centre of curvature, beyond the centre of curvature, or between the focal point and the pole, will cause the light rays to converge or diverge, creating a beam of light that is not parallel. Therefore, the correct option is "At the focal point".

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The wave transmitted in which the direction of vibration of the particles of a medium is perpendicular to the direction of travel of the wave is called a transverse wave. In a transverse wave, the oscillations of the particles of the medium occur at right angles to the direction of propagation of the wave. Examples of transverse waves include waves on a string, electromagnetic waves, and seismic S-waves.

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The nucleon number of an atom is the sum of its protons and neutrons. Therefore, the number of neutrons in an atom can be calculated by subtracting the number of protons (also called the atomic number) from the nucleon number. In this case, the nucleon number is given as 238 and the proton number (atomic number) is given as 92. Thus, the number of neutrons can be calculated as: Number of neutrons = nucleon number - proton number Number of neutrons = 238 - 92 Number of neutrons = 146 Therefore, the answer is 146.

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The absolute zero temperature is defined as the temperature at which thermal motion ceases. This is the lowest possible temperature that can be reached, and it is equivalent to zero Kelvin (0 K) or approximately -273.15 degrees Celsius (-459.67 degrees Fahrenheit). At this temperature, the particles that make up matter have the lowest possible energy and therefore stop moving completely. This is because temperature is a measure of the average kinetic energy of particles in a substance, and at absolute zero, the kinetic energy is zero. This concept is important in many areas of physics, such as thermodynamics and quantum mechanics, and it has practical applications in areas such as cryogenics, where temperatures approaching absolute zero are used to achieve superconductivity and other unique properties of matter.

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The power (P) of the lamp is given as 40 W and the voltage (V) is given as 240 V. Using Ohm's Law, we can calculate the resistance (R) of the filament of the lamp. Ohm's Law states that V = IR, where V is the voltage, I is the current, and R is the resistance. Using the formula for power, we know that P = IV. Substituting Ohm's Law into this equation gives P = (V²/R). Rearranging this formula gives R = V²/P. Substituting the given values gives R = (240²/40) = 1440 Ω. Therefore, the resistance of the filament of the lamp is 1440 Ω. Answer: 1440Ω

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Heat transfer by convection in a liquid is due to the translatory motion of the molecules of the liquid. When a liquid is heated, the molecules gain kinetic energy and move faster, which results in the formation of convection currents. These currents carry the heat energy from one point to another in the liquid. As the heated liquid rises, it transfers the heat to the surrounding cooler liquid molecules by collisions, causing the cooler liquid to become less dense and rise as well. This creates a continuous cycle of rising and sinking of the liquid, which transfers heat throughout the liquid. This process is known as convection, and it is the main mechanism of heat transfer in fluids like liquids and gases.

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Contact forces are forces that require physical contact between two objects to occur. Based on this definition, only forces of tension and friction are considered contact forces among the options given. Force of tension is a force that is transmitted through a string, rope, or cable when it is pulled tight by forces acting from opposite ends. It is a contact force because it requires physical contact between the string or cable and the objects pulling on it. Force of friction is a force that opposes motion between two surfaces in contact. It is a contact force because it requires physical contact between the two surfaces. Magnetic force and force of reaction are not contact forces because they do not require physical contact between two objects to occur. Magnetic force occurs between two magnetic objects, and force of reaction is the force exerted by a surface in response to an object pressing against it. Therefore, the correct answer is (b) I, II, and IV only.

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The half-life of a radioactive substance is the time taken for half of the radioactive nuclei to decay. After one half-life, half of the radioactive nuclei have decayed and only half remain. Similarly, after two half-lives, only a quarter of the original radioactive nuclei remain, and after three half-lives, only an eighth remain. In this question, the half-life is given as 20 hours, and the time elapsed is 80 hours, which is 4 times the half-life. Therefore, the number of half-lives that have passed is 4. So, after 4 half-lives, the fraction of the original radioactive nuclei remaining is (1/2) x (1/2) x (1/2) x (1/2) = 1/16. Therefore, the answer is: 1/16 of the original radioactive nuclei will remain after 80 hours.

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The reading on the spring - weighing machine will be greater than the actual weight of the girl. This is because when the lift starts to ascend, the girl and the weighing machine experience an upward force due to the acceleration of the lift. This force adds to the force of gravity acting on the girl and thus, the reading on the weighing machine increases. To find the reading on the weighing machine, we need to calculate the total force acting on the girl. Using Newton's second law, F=ma, the total force acting on the girl is the sum of the force of gravity and the force due to the lift's acceleration: F_total = (55 kg)(9.81 m/s^2) + (55 kg)(2 m/s^2) F_total = 676.05 N Therefore, the reading on the weighing machine will be: Reading = F_total = 676.05 N So the answer is 66 kg when converted to mass.

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The main difference between a crystalline and an amorphous solid is in their structure. A crystalline solid has a highly ordered, repeating pattern of molecules or atoms, which creates a well-defined shape and structure. This means that a crystalline solid has a regular geometric shape, with flat faces and sharp edges. On the other hand, an amorphous solid lacks this repeating pattern and does not have a well-defined shape. Instead, it has a more random arrangement of molecules or atoms, which creates a less ordered, less defined structure. This means that an amorphous solid has a more irregular shape, with rounded or curved surfaces. Examples of crystalline solids include diamonds, salt, and sugar, while examples of amorphous solids include glass, rubber, and plastic. In summary, the main difference between crystalline and amorphous solids is in their molecular or atomic arrangement, with crystalline solids having a highly ordered, repeating pattern and amorphous solids having a more random, less defined structure.

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Diffraction is not an evidence of the particle nature of matter. Diffraction is the bending of waves around an obstacle or through an opening. It is a property of waves, and it is not specific to the particle nature of matter. In fact, diffraction is often used to demonstrate the wave-like behavior of particles, such as electrons. On the other hand, the other options listed are evidences of the particle nature of matter: - Diffusion is the process of particles spreading out from an area of high concentration to an area of low concentration. It is a result of the movement of individual particles, and it is evidence of their particle-like nature. - Brownian motion is the random movement of particles in a fluid due to collisions with other particles. This motion is caused by the individual particles that make up the fluid, and it is evidence of their particle-like nature. - Crystal structure refers to the orderly arrangement of atoms or molecules in a solid. This structure is a result of the individual particles that make up the solid, and it is evidence of their particle-like nature. - Photoelectricity is the emission of electrons from a material when light shines on it. This phenomenon is evidence of the particle-like nature of light, which is made up of particles called photons. Therefore, the correct answer is "Diffraction".

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The force of gravitational attraction between two masses can be calculated using the formula F = G(m1m2)/r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers of mass. Using this formula, and plugging in the given values, we have: F = (7 x 10^-11 Nm^2/kg^2) x (10^6 kg) x (10^3 kg) / (1 m)^2 F = 7 x 10^-2 N Therefore, the force of attraction between the lead mass and the iron mass is 7 x 10^-2 N.

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The amount of heat given out by the iron can be calculated using the formula: Q = mcΔT Where Q is the heat energy transferred, m is the mass of the iron, c is the specific heat capacity of iron, and ΔT is the change in temperature of the iron. Substituting the given values, we have: Q = (0.05 kg) × (460 J/kg·K) × (85°C - 25°C) Q = 0.05 kg × 460 J/kg·K × 60°C Q = 1380 J Therefore, the amount of heat given out by the iron when it cools from 85°C to 25°C is 1380 J. Hence the correct option is (d) 1.38x10³J.

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The time of flight of an object in projectile motion can be calculated using the formula: t = 2u sin?/g where u is the initial velocity of the object, ? is the angle of projection, and g is the acceleration due to gravity. In this question, the object R leaves the platform with a horizontal velocity of 7m/s, which means its initial vertical velocity is 0. We can therefore use the formula: t = 2h/g where h is the height of the platform XY. We can rearrange this equation to solve for h: h = 1/2gt^2 We are given that it takes 0.3s for the object to fall freely from Y to P, so: h = 1/2 × 10 × (0.3)² = 0.45m Now, we can use the fact that the horizontal distance traveled by the object is given by: d = ut where d is the horizontal distance, u is the initial horizontal velocity, and t is the time of flight. Since there is no horizontal acceleration, the time of flight is the same as the time it takes for the object to fall from Y to P. Therefore: d = 7 × 0.3 = 2.1m Finally, we can use Pythagoras' theorem to find the distance PQ: PQ = ?(d² + h²) = ?(2.1² + 0.45²) ? 2.16m Therefore, the answer is 2.10m.

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The phenomenon responsible for water not dripping through an open silk umbrella is surface tension. Surface tension is the property of a liquid that causes its surface to behave like a stretched elastic membrane. The water droplets adhere to the silk fabric of the umbrella and are held in place by the cohesive forces between water molecules. The umbrella's fabric has a network of fibers, which creates capillary action that results in an upward force that balances the weight of the water, making it difficult for the water to fall through the umbrella. Only when the inside of the umbrella is touched and the tension is disrupted, the water droplets fall through the fabric.

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When an object is placed in front of a concave mirror, the mirror forms an image of the object. The image can either be real or virtual, and it can be magnified or diminished depending on the location of the object and the focal length of the mirror. In this case, the concave mirror has a radius of curvature of 20cm, which means its focal length is half of that or 10cm. The pin is placed at a distance of 15cm from the pole of the mirror, which means it is located beyond the focal point. Since the object is placed beyond the focal point, the image formed will be real and inverted. The magnification of the image can be calculated using the formula: magnification = height of image / height of object = -v/u where v is the distance of the image from the mirror, u is the distance of the object from the mirror, and the negative sign indicates that the image is inverted. Using the mirror formula (1/f = 1/v + 1/u), we can find the distance of the image from the mirror: 1/f = 1/v + 1/u 1/10 = 1/v + 1/15 v = 30cm Now, we can use the magnification formula to find the magnification of the image: magnification = height of image / height of object = -v/u magnification = -30/15 magnification = -2.00 The negative sign indicates that the image is inverted, and the magnification is 2.00. Therefore, the answer is 2.00.

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The distance of the bat from the reflector can be calculated using the speed of sound and the time taken for the echo to return. The speed of sound is given as 1650.0ms^-1. When the bat emits a sound wave, it travels to the reflector and back, covering a total distance equal to twice the distance from the bat to the reflector. The time taken for this round trip is 0.15s (as given in the question). Using the equation: Distance = Speed x Time We can calculate the distance of the bat from the reflector as: Distance = (Speed of sound x Time taken)/2 Distance = (1650.0ms^-1 x 0.15s)/2 Distance = 123.75m Therefore, the correct answer is (d) 123.75m.

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To determine the reading of the micrometer screw gauge, we need to add the main scale reading to the circular scale reading. From the image, we can see that the main scale reading is 3.0 mm, which is the distance shown between the left end of the sleeve and the central zero mark on the thimble. To find the circular scale reading, we count the number of divisions on the circular scale that line up with the main scale. We see that the 20th line on the circular scale is aligned with the main scale. Since each division on the circular scale represents 0.01 mm, we can calculate the circular scale reading as 20 x 0.01 mm = 0.20 mm. Therefore, the total reading is 3.0 mm + 0.20 mm = 3.20 mm. None of the answer options match our calculated value. The closest option is 3.17 mm.

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To solve this problem, we need to use the formula: Q = m c ΔT where Q is the heat energy required to heat the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. The specific heat capacity of water is 4.18 J/g°C, and since we have 3 kg of water, the mass of water is 3000 g. So, Q = 3000 g × 4.18 J/g°C × (88°C - 28°C) = 376440 J Now, we can use the formula for electrical energy: E = V I t where E is the electrical energy, V is the voltage, I is the current, and t is the time. Rearranging the formula to solve for t, we get: t = E / (V I) Substituting the given values, we get: t = 376440 J / (220 V × 6 A) = 570 s Therefore, the time it will take to heat 3 kg of water from 28°C to 88°C in an electric taking a current of 6 A from an e.m.f. source of 220V is 570 seconds or 9.5 minutes. So, the correct answer is 570s.

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The period of a pendulum is the time it takes for one complete oscillation, which is the time it takes for the pendulum to swing back and forth once. In this question, we are given that the simple pendulum makes 50 oscillations in one minute. To find the period of oscillation, we need to divide the total time taken by the number of oscillations. We can convert one minute to seconds by multiplying it by 60, since there are 60 seconds in one minute. So, the time taken for 50 oscillations is: Time = 1 minute/50 oscillations = 60 seconds/50 oscillations = 1.2 seconds/oscillation Therefore, the period of oscillation of the pendulum is 1.2 seconds. Therefore, the correct option is (d) 1.20s.

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The density of water decreases as its temperature increases from 0^oC to 4^oC, after which it increases. Therefore, as the temperature of the small quantity of water in a closed container is gradually increased from 0^oC to 40^oC, its density will first decrease until it reaches a minimum value at 4^oC, and then it will increase. Therefore, the correct answer is (A) "increases for a while and then decreases".

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The work done (W) in transferring charge (q) between two points in an electric field is given by the formula: W = qV Where V is the potential difference between the two points. In this question, the work done is 30 J and the charge transferred is 5 mC (or 5 x 10^-3 C). Therefore, we can rearrange the formula to find the potential difference: V = W/q = 30 J / 5 x 10^-3 C = 6,000 V So the potential difference between points B and A is 6,000 volts. Therefore, the correct option is: - 6.0 x 10^3 V

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To solve this problem, we need to use the conservation of mechanical energy of the pendulum. At the highest point P, the pendulum is momentarily at rest, and all its kinetic energy has been converted to potential energy. Therefore, the sum of kinetic and potential energies at P is equal to the sum of kinetic and potential energies at O. At position O, the pendulum has only kinetic energy, which is given by the formula 1/2 * m * v^2, where m is the mass of the pendulum and v is its velocity. We are not given the mass of the pendulum, but we can assume that it is constant throughout the motion. Therefore, the kinetic energy at O is 1/2 * m * (2)^2 = 2m. At position P, the pendulum has only potential energy, which is given by the formula m * g * h, where h is the height of P above O. Therefore, the potential energy at P is m * g * h. Since the total mechanical energy of the pendulum is conserved, we can equate the kinetic energy at O to the potential energy at P: 2m = m * g * h Simplifying this equation, we get: h = 2 / g = 2 / 10 = 0.2m Therefore, the height of P above O is 0.2m. Answer option (D) is correct.

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Particles that conduct electricity through salty water are called ions. When salt is dissolved in water, it dissociates into positively charged ions (such as sodium ions) and negatively charged ions (such as chloride ions). These ions are able to move freely in the water, carrying an electric charge and allowing an electric current to flow through the solution. In contrast, atoms and molecules do not conduct electricity in water because they do not have a net electric charge and are not able to move freely. Electrons and neutrons are subatomic particles that can be found in atoms, but they are not able to move freely in water and therefore cannot conduct electricity in this medium. Therefore, the correct option is "Ions".

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An induction coil is a device that converts high voltage, low current power to low voltage, high current power through electromagnetic induction. When an alternating current passes through the primary coil, a changing magnetic field is produced which induces an alternating current in the secondary coil. The back e.m.f. generated in the primary coil of an induction coil opposes the flow of current in the coil, causing a decrease in the current. This effect can be reduced by using a capacitor in the circuit, which stores the charge and releases it when the current drops. This maintains the flow of current and reduces the effect of the back e.m.f. Therefore, option I is correct. The make and break contact in the circuit also helps to reduce the effect of the back e.m.f. The contact repeatedly breaks the circuit, causing the current to stop flowing momentarily, which interrupts the magnetic field and reduces the back e.m.f. Therefore, option II is also correct. The ratio of turns in the secondary to that in the primary does not affect the back e.m.f. generated in the primary coil. It only affects the voltage induced in the secondary coil. Therefore, option III is incorrect. Hence, the correct answer is option C, I and II only.

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To calculate the current in the circuit, we need to use Ohm's law and Kirchhoff's circuit laws. First, we can calculate the total resistance of the circuit by adding the resistance of the ammeter and the resistor in series: R = 0.59 Ω + 7.09 Ω = 7.68 Ω Using Kirchhoff's circuit laws, we know that the current I through the circuit is equal to the e.m.f. E of the cell divided by the total resistance R of the circuit plus the internal resistance r of the cell: I = E / (R + r) Substituting the given values, we get: I = 1.5 V / (7.68 Ω + 2.59 Ω) = 0.15 A Therefore, the current in the circuit is 0.15 A. So, the answer is 0.15 A.

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The electric field intensity (E) at a point in space is defined as the electric force (F) per unit charge (q) at that point. Therefore, we can use the formula: E = F/q Substituting the given values, we get: E = 40 N / 1.0 x 10^-5 C = 4.00 x 10⁶ N/C Thus, the electric field intensity at the given point in space is 4.00 x 10⁶ N/C. Therefore, the correct option is: 4.00 x 10⁶NC^-1.

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The body P is in equilibrium, which means that the net force acting on it is zero. The weight of P acts downwards and is balanced by the tension in the string acting upwards. The weight of P is given by: Weight of P = mP * g = 20 kg * 10 ms^-2 = 200 N Since the net force acting on P is zero, the tension in the string is equal to the weight of P. Therefore: Tension in string = Weight of P = 200 N So, option (E) is the correct answer.

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The problem involves a trapped gas at a fixed volume in a closed-end tube, and we are asked to find the temperature at which the length of the gas column will increase to a specified value. This can be solved using the relationship between the volume of a gas and its temperature, known as Charles's Law. According to Charles's Law, the volume of a gas is directly proportional to its temperature at constant pressure. Mathematically, this can be expressed as: V1/T1 = V2/T2 where V1 and T1 are the initial volume and temperature, respectively, and V2 and T2 are the final volume and temperature, respectively. In this problem, the volume of the gas is constant since it is trapped by the mercury pellet, so we can simplify the equation to: T1/T2 = L1/L2 where L1 and L2 are the initial and final lengths of the gas column, respectively. We are given that the initial length of the gas column is 1.0 m at a temperature of 27°C (which can be converted to 300.15 K). We are asked to find the temperature at which the length of the gas column will increase to 1.20 m. Using the above equation, we can solve for T2: T2 = (L2/L1) x T1 T2 = (1.20 m / 1.0 m) x 300.15 K T2 = 360.18 K We can convert this temperature to degrees Celsius by subtracting 273.15: T2 = 87.03°C Therefore, the correct option is (d) 87.0°C.

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(a) Obtaining quantities from a velocity-time graph

On a velocity-time graph the velocity is plotted on the vertical axis and time on the horizontal axis. A typical graph rises from A to B, stays level from B to C, then falls from C to D as shown below.

(i) Acceleration is obtained from the gradient of the rising portion AB. If the velocity increases from \(v_1\) to \(v_2\) as time changes from \(t_1\) to \(t_2\) along AB, then \[ a = \text{gradient of AB} = \frac{v_2 - v_1}{t_2 - t_1} \quad (\text{a positive slope}). \]

(ii) Retardation (deceleration) is obtained from the gradient of the falling portion CD. Here the velocity decreases with time, so the slope is negative: \[ \text{retardation} = -\,(\text{gradient of CD}) = \frac{v_C - v_D}{t_D - t_C}. \]

(iii) Total distance covered is the total area enclosed between the graph line and the time axis (the shaded region O-A-B-C-D). It is found by splitting the region into triangles and a rectangle (or one trapezium) and adding the areas.

(b) Deriving \(s = ut + \tfrac{1}{2}at^{2}\)

The displacement equals the area under the velocity-time graph, which for uniform acceleration is a trapezium of parallel sides \(u\) (initial velocity) and \(v\) (final velocity) and width \(t\): \[ s = \text{average velocity} \times t = \left(\frac{u + v}{2}\right)t. \tag{1} \] For uniform acceleration \(a\), the final velocity is \[ v = u + at. \tag{2} \] Substituting (2) into (1): \[ s = \left(\frac{u + (u + at)}{2}\right)t = \left(\frac{2u + at}{2}\right)t \] \[ \boxed{\,s = ut + \tfrac{1}{2}at^{2}\,}. \]

(c) Uniform deceleration problem

Along the straight line the velocities are: at P, \(u = 40\,\text{ms}^{-1}\); at Q, \(20\,\text{ms}^{-1}\); at R the particle is at rest (\(0\,\text{ms}^{-1}\)); and \(QR = 50\,\text{m}\).

Find the deceleration using the motion from Q to R, with \(v^{2} = u^{2} - 2as\): \[ 0 = 20^{2} - 2a(50) \;\Rightarrow\; 400 = 100a \;\Rightarrow\; a = 4\,\text{ms}^{-2}\ (\text{deceleration}). \]

(i) Distance PQ (from P to Q): \[ 20^{2} = 40^{2} - 2(4)(PQ) \;\Rightarrow\; 400 = 1600 - 8\,PQ \;\Rightarrow\; PQ = \frac{1200}{8} = 150\,\text{m}. \]

(ii) Time taken to cover PQ, using \(v = u - at\) from P to Q: \[ 20 = 40 - 4t \;\Rightarrow\; 4t = 20 \;\Rightarrow\; t = 5\,\text{s}. \]

(iii) Time taken to cover PR (from P to rest at R): \[ 0 = 40 - 4t \;\Rightarrow\; 4t = 40 \;\Rightarrow\; t = 10\,\text{s}. \]

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(a) Boiling point

The boiling point of a liquid is the constant temperature at which the liquid changes to vapour throughout its bulk, when its saturated vapour pressure equals the external (atmospheric) pressure.

(b) Experiment to determine the boiling point of a small quantity of liquid

1. Put the small quantity of liquid in a narrow ignition tube.
2. Attach a thermometer to the tube with a rubber band so the bulb is level with the liquid.
3. Immerse the tube and thermometer in a beaker of water (a heating bath) clamped on a stand.
4. Heat the bath steadily while stirring, and observe the thermometer.
5. The temperature at which the liquid boils steadily (bubbles rise continuously and temperature stays constant) is read and recorded as the boiling point.

(c) Mass of water that boils away

Heat lost by copper cooling from 950°C to 100°C:

\[ Q_{Cu} = m_{Cu} c_{Cu}\,\Delta\theta = 0.300 \times 4.0\times10^{2} \times (950-100) = 0.300\times400\times850 = 102000\,\text{J} \]

Heat used to raise the water from 25°C to 100°C:

\[ Q_{w} = 0.250 \times 4.2\times10^{3} \times (100-25) = 0.250\times4200\times75 = 78750\,\text{J} \]

Heat left for vaporization:

\[ Q_{v} = 102000 - 78750 = 23250\,\text{J} \]

Mass boiled away:

\[ m = \frac{Q_{v}}{L} = \frac{23250}{2.26\times10^{6}} = 1.03\times10^{-2}\,\text{kg} \approx 10.3\,\text{g} \]

(d) Four other effects of heat (besides expansion)

• Change of state (melting, boiling, evaporation).
• Rise in temperature of the substance.
• Chemical change (e.g. decomposition or burning).
• Change in electrical resistance (or thermionic/thermoelectric effects).

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None

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(a) Capacitance of a capacitor

The capacitance of a capacitor is the ratio of the magnitude of the charge \(Q\) on either plate to the potential difference \(V\) between the plates:

\[ C = \frac{Q}{V} \]

where \(Q\) is the charge (coulomb) and \(V\) is the p.d. (volt). Its SI unit is the farad (F).

(b) Factors affecting the capacitance of a parallel-plate capacitor

1. The common (overlapping) area \(A\) of the plates, with \(C \propto A\).
2. The distance of separation \(d\) between the plates, with \(C \propto \dfrac{1}{d}\).
3. The permittivity \(\varepsilon\) (nature) of the dielectric medium between the plates, with \(C \propto \varepsilon\).

These combine as \(C = \dfrac{\varepsilon A}{d}\).

(c) Energy \(W\) stored in a charged capacitor

When the capacitor already holds a charge \(q\), the p.d. across it is \(v = \dfrac{q}{C}\). To move a further small charge \(dq\) onto the plates the work done is

\[ dW = v\,dq = \frac{q}{C}\,dq \]

The total work to charge the capacitor from \(0\) to the final charge \(Q\) is therefore

\[ W = \int_{0}^{Q} \frac{q}{C}\,dq = \frac{1}{C}\cdot\frac{Q^{2}}{2} = \frac{Q^{2}}{2C} \]

Using \(Q = CV\), the same result may be written as

\[ W = \frac{Q^{2}}{2C} = \frac{1}{2}QV = \frac{1}{2}CV^{2} \]

(d) Two capacitors 4\(\mu\text{F}\) and 6\(\mu\text{F}\) in series across 100 V

Circuit diagram: the two capacitors are joined end to end (in series) in a single loop with the 100 V d.c. supply.

Effective (combined) capacitance of capacitors in series:

\[ \frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{1}{4} + \frac{1}{6} = \frac{3+2}{12} = \frac{5}{12} \] \[ \Rightarrow C = \frac{12}{5} = 2.4\,\mu\text{F} \]

(i) Charge on either plate of each capacitor. In a series circuit the charge is the same on every capacitor and equals the charge supplied to the combination:

\[ Q = CV = 2.4\times10^{-6} \times 100 = 2.4\times10^{-4}\,\text{C} = 240\,\mu\text{C} \]

Hence each capacitor carries \(Q = 240\,\mu\text{C}\) on either plate.

(ii) P.d. across each capacitor.

\[ V_{4} = \frac{Q}{C_{4}} = \frac{2.4\times10^{-4}}{4\times10^{-6}} = 60\,\text{V} \] \[ V_{6} = \frac{Q}{C_{6}} = \frac{2.4\times10^{-4}}{6\times10^{-6}} = 40\,\text{V} \]

Check: \(V_{4} + V_{6} = 60 + 40 = 100\,\text{V}\), which equals the supply voltage.

(iii) Energy of the combined capacitors.

\[ W = \frac{Q^{2}}{2C} = \frac{(2.4\times10^{-4})^{2}}{2 \times 2.4\times10^{-6}} = \frac{5.76\times10^{-8}}{4.8\times10^{-6}} = 1.2\times10^{-2}\,\text{J} \]

Equivalently \(W = \tfrac{1}{2}QV = \tfrac{1}{2}(2.4\times10^{-4})(100) = 0.012\,\text{J}\).

Therefore the energy stored in the combination is \(W = 1.2\times10^{-2}\,\text{J}\).