An electron of mass m and charge \(\Box\) enters a uniform electric field between two metal plates P and q separated by a distance d. P is maintained at a p...

An electron of mass m and charge \(\Box\) enters a uniform electric field between two metal plates P and q separated by a distance d. P is maintained at a potential V while Q is earthed. Determine an expression for the magnitude of the acceleration of electron through the field

Answer Details

The electric field between the two metal plates exerts a force on the electron which accelerates it. According to Newton's second law, the magnitude of the acceleration (a) of an object is proportional to the net force (F) acting on it, and inversely proportional to its mass (m), i.e., a = F/m. In this case, the net force acting on the electron is the electric force (Fe) given by Fe = qE, where q is the charge of the electron and E is the magnitude of the electric field between the plates. Since the electron has a negative charge, the direction of the force is opposite to the direction of the field, so the electron is accelerated towards the negative plate. The electric field between the plates is given by E = V/d, where V is the potential difference between the plates and d is the distance between them. Therefore, the force on the electron is given by Fe = q(V/d). Substituting this expression for Fe into the equation for acceleration, we get: a = Fe/m = q(V/d) / m Since the charge of an electron is -e (where e is the elementary charge), we can substitute -e for q to get: a = (-e)(V/d) / m Simplifying this expression, we get: a = - (eV/md) So the magnitude of the acceleration of the electron is given by: a = (eV/md) Therefore, the correct option is (a) \(\frac{eV}{md}\).