Question

In moving from A to B along an electric field line, the electric field does 1.28 * 10-18 J
of work on an electron. If 01.02 are equipotential surfaces, then potential difference
VE-Vo is equal to
BI​

Answer 1

ANSWER: hope it helps you..

Answer 2

Given info : In moving from A to B along an electric field line, the electric field does 1.28 × 10^-18 J of work on an electron.

To find : If Φ₁ and Φ₂ are equipotential surfaces , then potential difference $V_C-V_A$ is ..

solution : see diagram, it is clear that point B and C are located at the same equipotential surface, Φ₂.

so, potential at B = potential at C

⇒$V_B=V_C$ ....(1)

Given, work done to an electron from A to B is 1.28 × 10^-18 J

so, W = e$(V_A-V_B)$

where e is charge of electron.

⇒1.28 × 10^-18 = -1.6 × 10^-19$(V_A-V_B)$

⇒-8 = $(V_A-V_B)$

from equation (1) we get,

$(V_A-V_C)$ = -8 volts

⇒$(V_C-V_A)$ =8 volts

Therefore the potential difference $(V_C-V_A)$ is 8 volts.