V0 is the potential at the origin in an electric field- E-Exi-Eyj- The potential at the point x-y is-

The correct option is A V_0 xE_x+yE_y We know V(x_1) V(x_2)=∫_x_1^x_2 E d x. Now potential change in x direction V_x(x) V_x(o)=∫_x^o E_x dx= E_x x ...

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Byju's Answer

Standard XII

Physics

Electric Field

V0 is the pot...

Question

$V_{0}$ is the potential at the origin in an electric field. $\to E = E_{x} \to i - E_{y} \to j$ . The potential at the point (x,y) is:

V_{0} - x E_{x} + y E_{y}

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V_{0} + x E_{x} + y E_{y}

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x E_{x} + y E_{y} - V_{0}

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\sqrt{x^{2} + y^{2}} \sqrt{E^{2} x + E^{2} y - V_{0}}

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Solution

The correct option is A $V_{0} - x E_{x} + y E_{y}$
We know
$V (x_{1}) - V (x_{2}) = \int_{x_{1}}^{x_{2}} E d x .$
Now potential change in x direction
$V_{x} (x) - V_{x} (o) = \int_{x}^{o} E_{x} d x = - E_{x} x .$
$V_{x} (x) = V_{x} (o) - E_{x} x$
Potential change in y direction
$V_{y} (y) - V_{y} (o) = \int_{y}^{o} (- E_{y}) d y = + E_{y} y$
$V_{y} (y) = V_{y} (o) + E_{y} y$
Now total potential at (x,y)
$= V_{x} + V_{y}$
$= V_{x} (o) - E_{x} x + V_{y} (o) + E_{y} y$
$= V_{o} - x E_{x} + y E_{y}$

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किसी त्रिविम क्षेत्र में, माना एक विद्युत क्षेत्र

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उपस्थित है, जहाँ x मीटर में हैं। विभवान्तर V_A – V₀ का मान क्या होगा, जहाँ V₀ मूलबिंदु का विभव है तथा V_A, बिंदु x = 2 m पर विभव है?

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V0 is the potential at the origin in an electric field. →E=Ex→i−Ey→j. The potential at the point (x,y) is:

The correct option is A V0−xEx+yEyWe knowV(x1)−V(x2)=∫x2x1Edx.Now potential change in x directionVx(x)−Vx(o)=∫oxExdx=−Exx.Vx(x)=Vx(o)−ExxPotential change in y directionVy(y)−Vy(o)=∫oy(−Ey)dy=+EyyVy(y)=Vy(o)+EyyNow total potential at (x,y)=Vx+Vy=Vx(o)−Exx+Vy(o)+Eyy=Vo−xEx+yEy

$V_{0}$ is the potential at the origin in an electric field. $\to E = E_{x} \to i - E_{y} \to j$ . The potential at the point (x,y) is: