Answer to Question #229639 in Electric Circuits for spence

Question

A point charge of 5 µC is located at point A between two horizontally parallel plates, which are 40 cm apart. The potential of the upper plate is 1000 V and the lower plate is at zero. The point charge then moves 30cm downwards from point A to point B. Given that the force when the point charge is at A is 0.0125N:

a) calculate the force on the charge when it is moved to position B

b) Calculate the energy that is expended in moving the charge from point A to point B

Answer 1

Explanations & Calculations

• The force imposed on a charge in an electric filed is given by $\small F= Eq$. Then we can calculate the electric filed at point A by considering the given data,

$\qquad\qquad \begin{aligned} \small E&=\small \frac{0.0125\,N}{5\times10^{-6}\,C}\\ &=\small 2500\,NC^{-1}=2500\,Vm^{-1} \end{aligned}$

• The uniform electric field spreaded over a large area which is also the same as that between large parallel plates is given $\small E =\large \frac{V}{d}$. Checking whether this applies here also we get,

$\qquad\qquad \begin{aligned} \small E&=\small \frac{(1000-0)V}{0.4\,m}\\ &=\small 2500\,Vm^{-1} \end{aligned}$

• Results coincide. Therefore, the electric field is uniform all over the region.

a)

• Therefore, according to the equation $\small F=Eq$, the force on the charge at B is the same as that at A: the constant 0.0125N.

b)

• The energy expended is,

$\qquad\qquad \begin{aligned} \small W&=\small F\times d\\ &=\small 0.0125\,N\times0.3\,m\\ &=\small \bold{3.75\,mJ} \end{aligned}$