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The electric potential at a space point ...

The electric potential at a space point P (X, Y, Z) is given as `V= x^(2)+y^(2)+z^(2)`. The modulus of the electric field at that point is proportional to

A

(a)`V^((1)/(2))`

B

(b)V

C

(c)`V^((3)/(2))`

D

(d)`V^(2)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the relationship between the electric potential \( V \) and the electric field \( E \) at the point \( P(X, Y, Z) \). The electric potential is given as: \[ V = x^2 + y^2 + z^2 \] ### Step 1: Understanding the Electric Field The electric field \( E \) is related to the electric potential \( V \) by the equation: \[ \vec{E} = -\nabla V \] where \( \nabla V \) is the gradient of the potential \( V \). ### Step 2: Calculating the Gradient of \( V \) The gradient in three dimensions is given by: \[ \nabla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right) \] Now, we calculate each of these partial derivatives. 1. **Partial derivative with respect to \( x \)**: \[ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(x^2 + y^2 + z^2) = 2x \] 2. **Partial derivative with respect to \( y \)**: \[ \frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(x^2 + y^2 + z^2) = 2y \] 3. **Partial derivative with respect to \( z \)**: \[ \frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(x^2 + y^2 + z^2) = 2z \] Thus, the gradient of \( V \) is: \[ \nabla V = (2x, 2y, 2z) \] ### Step 3: Finding the Electric Field Now, substituting the gradient into the equation for the electric field: \[ \vec{E} = -\nabla V = (-2x, -2y, -2z) \] ### Step 4: Calculating the Magnitude of the Electric Field The magnitude of the electric field \( E \) is given by: \[ E = |\vec{E}| = \sqrt{(-2x)^2 + (-2y)^2 + (-2z)^2} \] Calculating this gives: \[ E = \sqrt{4x^2 + 4y^2 + 4z^2} = 2\sqrt{x^2 + y^2 + z^2} \] ### Step 5: Relating \( E \) to \( V \) Since we know that: \[ V = x^2 + y^2 + z^2 \] We can express the electric field in terms of \( V \): \[ E = 2\sqrt{V} \] ### Conclusion The modulus of the electric field \( E \) at that point is proportional to \( \sqrt{V} \). Thus, the answer is: \[ \text{The modulus of the electric field at that point is proportional to } V^{1/2}. \]
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