(a) (i) Explain (I) electric potential; (II) electric potential energy
(ii) State the SI unit of each of the term in (a)(i) above
(b) An isolated electrically charged sphere of radius, r, and charge, Q, is supported on an insulator in air of permitivity, \(\varepsilon_o\). Write down;
(i) an expression for the electric field intensity on the surface of the sphere;
(ii) an expression for the electric potential at the surface of the sphere;
(iii) a relationship between the electric field intensity and the electric potential at the surface of the sphere
(c) The plates of a parallel plate capacitor, 5.0 x 10\(^{-3}\) m apart are maintained at a potential difference of 5.0 x 10\(^{4}\) V. Calculate the magnitude f the
(i) electric field intensity between the plates
(ii) force on the electron
(iii) acceleration of the elctron
[electronic charge = 1.60 x 10\(^{-19}\)C, mass of electron = 9.1 \times 10\(^{-31}\) kg]
(a)(i)(I) Electric potential at a point in an electric field is the work done in bringing a unit positive charge from infinity to that point.
(a)(i)(II) Electric potential energy of a charge at a point is the work done in bringing that charge from infinity to the point in the electric field.
(a)(ii) SI units: electric potential is measured in the volt (V); electric potential energy is measured in the joule (J).
(b)(i) Electric field intensity at the surface of the charged sphere: \[ E = \dfrac{Q}{4\pi \varepsilon_o r^2}. \]
(b)(ii) Electric potential at the surface of the sphere: \[ V = \dfrac{Q}{4\pi \varepsilon_o r}. \]
(b)(iii) Relationship between them: \[ E = \dfrac{V}{r}\quad(\text{i.e. } V = E r). \]
(c)(i) Electric field intensity between the plates: \[ E = \dfrac{V}{d} = \dfrac{5.0 \times 10^{4}}{5.0 \times 10^{-3}} = 1.0 \times 10^{7}\,\text{V m}^{-1}. \]
(c)(ii) Force on the electron: \[ F = eE = 1.60 \times 10^{-19} \times 1.0 \times 10^{7} = 1.6 \times 10^{-12}\,\text{N}. \]
(c)(iii) Acceleration of the electron: \[ a = \dfrac{F}{m} = \dfrac{1.6 \times 10^{-12}}{9.1 \times 10^{-31}} = 1.76 \times 10^{18}\,\text{m s}^{-2}. \]
(a)(i)(I) Electric potential at a point in an electric field is the work done in bringing a unit positive charge from infinity to that point.
(a)(i)(II) Electric potential energy of a charge at a point is the work done in bringing that charge from infinity to the point in the electric field.
(a)(ii) SI units: electric potential is measured in the volt (V); electric potential energy is measured in the joule (J).
(b)(i) Electric field intensity at the surface of the charged sphere: \[ E = \dfrac{Q}{4\pi \varepsilon_o r^2}. \]
(b)(ii) Electric potential at the surface of the sphere: \[ V = \dfrac{Q}{4\pi \varepsilon_o r}. \]
(b)(iii) Relationship between them: \[ E = \dfrac{V}{r}\quad(\text{i.e. } V = E r). \]
(c)(i) Electric field intensity between the plates: \[ E = \dfrac{V}{d} = \dfrac{5.0 \times 10^{4}}{5.0 \times 10^{-3}} = 1.0 \times 10^{7}\,\text{V m}^{-1}. \]
(c)(ii) Force on the electron: \[ F = eE = 1.60 \times 10^{-19} \times 1.0 \times 10^{7} = 1.6 \times 10^{-12}\,\text{N}. \]
(c)(iii) Acceleration of the electron: \[ a = \dfrac{F}{m} = \dfrac{1.6 \times 10^{-12}}{9.1 \times 10^{-31}} = 1.76 \times 10^{18}\,\text{m s}^{-2}. \]
