Question

In the uniform electric field shown in figure, ${V_A} - {V_C}$ is

A) $10V$
B) $20V$
C) $40V$
D) $5V$

Answer 1

Hint- Potential difference between two points is defined as the total amount of work done in moving the positive charge from one point to another. The electric field is also due to the charge present on the particular area. It is measured in volt per metre. And it can also be said as the electric force.

Complete step by step answer:
(i) As we know both the potential difference and the electric field is due to the charge present on the particular area, the relation between both the potential difference and the electric field is, ${V_A} - {V_C} = E \times d$
(ii) From the diagram, we can understand the electric field of $20V{m^{ - 1}}$in the direction of A to B.
(iii) The plane AB and DC is acting in the uniform electric field. From A to B, we can say that the potential at A is higher than B as the electric field is acting in the direction from A to B.
(iv)A potential at D is greater than the potential at C. And the potential at D and the potential at C are equal to the potential at A and potential at B respectively. This is because both planes are in the same electric field. And they are called as equipotential surfaces.
(v) Therefore the potential between A and B and the potential between A and C are equal. Hence, ${V_A} - {V_C} = 20 \times 1$
$\therefore {V_A} - {V_C} = 20V$

(vi)Therefore the correct option is B.

Note: If the potential difference at two surfaces is equal, those two surfaces are equipotential surfaces. And the difference between the work-done on two surfaces is zero. The unit of potential difference is Volt. One volt is defined as the one joule of work done in moving one unit of positive charge from one place to another.