The potential V is varying with x and y as V - 12y2-4x volt- The field at x - 1 m - y - 1 m- is -

The correct option is C 2î ĵ Vm^ 1 #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; ...

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Standard XII

Physics

Electric Field Due to Charge Distributions - Approach

The potential...

Question

The potential $V$ is varying with x and y as $V = \frac{1}{2} (y^{2} - 4 x)$ volt. The field at $x = 1 m, y = 1 m$ , is :

2^i +^j V m^{- 1}

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- 2^i +^j V m^{- 1}

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2^i -^j V m^{- 1}

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- 2^i + 2^j V m^{- 1}

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Solution

The correct option is C $2^i -^j V m^{- 1}$

$E_{x} = - \frac{d V}{d x} = - \frac{1}{2} [- 4] = 2$

$E_{y} = - \frac{d V}{d y} = - \frac{1}{2} [2 y] = - y = - 1$

$∴ \to E = E_{x}^i + E_{y}^j = 2^i - 1^j$

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The potential V is varying with x and y as V=12(y2−4x) volt. The field at x=1m,y=1m, is :

The correct option is C 2^i−^j Vm−1 Ex=−dVdx=−12[−4]=2 Ey=−dVdy=−12[2y]=−y=−1 ∴→E=Ex^i+Ey^j=2^i−1^j

The potential $V$ is varying with x and y as $V = \frac{1}{2} (y^{2} - 4 x)$ volt. The field at $x = 1 m, y = 1 m$ , is :

The correct option is C $2^i -^j V m^{- 1}$

$E_{x} = - \frac{d V}{d x} = - \frac{1}{2} [- 4] = 2$

$E_{y} = - \frac{d V}{d y} = - \frac{1}{2} [2 y] = - y = - 1$

$∴ \to E = E_{x}^i + E_{y}^j = 2^i - 1^j$