Two large plane parallel conducting plates are kept 10 cm apart as shown in figure. The potential difference between them is V . The potential difference between the points A and B (shown in the figure) is :

Solution

Physics sir · 4 steps

Only the distance along the field counts; the 3 cm side runs parallel to the plates and adds nothing.

1. Uniform field. Between large parallel plates the field is uniform and perpendicular to the plates, so E = V/d = V/(10 cm). Here the plates are vertical, so E points horizontally.
2. A to C. AC is vertical, parallel to the plates and perpendicular to E. A and C lie on the same equipotential, so V_A = V_C.
3. C to B. CB lies along the field. From the right triangle, CB = √(5² - 3²) = 4 cm.

   V_(AB) = V_(CB) = E × CB = V/(10 cm) × 4 cm = 2V/5

4. So. The potential difference between A and B is 2/5V, option (d).

The idea to keep. In a uniform field, Δ V = E × (displacement component along E). Break any slanted path into a part along the field and a part across it. The part across it costs nothing.

Answer from the NTA final answer key. Solution by PrepWiser.