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The longer side of a rectangle is twice ...

The longer side of a rectangle is twice the length of its shorter side. A charge q is kept at one vertex. The maximum electric potential due to that charge at any other vertex is V, then the minimum electric potential at any other vertex will be

A

2V

B

`sqrt3V`

C

`V//sqrt5`

D

`sqrt5V`

Text Solution

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The correct Answer is:
To solve the problem step by step, we will analyze the rectangle and the electric potential due to the charge \( q \) at different vertices. ### Step 1: Define the rectangle dimensions Let the length of the shorter side of the rectangle be \( L \). Therefore, the longer side will be \( 2L \). ### Step 2: Identify the vertices Assume the rectangle has vertices at: - \( A(0, 0) \) (where the charge \( q \) is located) - \( B(L, 0) \) - \( C(L, 2L) \) - \( D(0, 2L) \) ### Step 3: Calculate distances from charge \( q \) to other vertices We need to find the distances from charge \( q \) at vertex \( A \) to the other vertices: - Distance to vertex \( B \): \( r_{AB} = L \) - Distance to vertex \( C \): \( r_{AC} = \sqrt{L^2 + (2L)^2} = \sqrt{L^2 + 4L^2} = \sqrt{5L^2} = L\sqrt{5} \) - Distance to vertex \( D \): \( r_{AD} = 2L \) ### Step 4: Calculate electric potential at each vertex The electric potential \( V \) due to a point charge \( q \) at a distance \( r \) is given by: \[ V = \frac{kq}{r} \] where \( k = \frac{1}{4\pi \epsilon_0} \). 1. **At vertex \( B \)**: \[ V_B = \frac{kq}{L} \] 2. **At vertex \( C \)**: \[ V_C = \frac{kq}{L\sqrt{5}} \] 3. **At vertex \( D \)**: \[ V_D = \frac{kq}{2L} \] ### Step 5: Identify maximum and minimum potentials From the calculations: - The maximum potential \( V_{max} \) occurs at vertex \( B \): \[ V_{max} = \frac{kq}{L} \] - The minimum potential \( V_{min} \) occurs at vertex \( C \): \[ V_{min} = \frac{kq}{L\sqrt{5}} \] ### Step 6: Relate \( V_{min} \) to \( V \) Given that the maximum electric potential at any other vertex is \( V \): \[ V = \frac{kq}{L} \] Thus, we can express the minimum potential in terms of \( V \): \[ V_{min} = \frac{V}{\sqrt{5}} \] ### Final Answer The minimum electric potential at any other vertex is: \[ V_{min} = \frac{V}{\sqrt{5}} \] ---

To solve the problem step by step, we will analyze the rectangle and the electric potential due to the charge \( q \) at different vertices. ### Step 1: Define the rectangle dimensions Let the length of the shorter side of the rectangle be \( L \). Therefore, the longer side will be \( 2L \). ### Step 2: Identify the vertices Assume the rectangle has vertices at: - \( A(0, 0) \) (where the charge \( q \) is located) ...
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