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I'm a physics tutor. This is the question from a High school book.

The equation of an equipotential line in an electric field is y = 2x, then the electric field strength vector at (1, 2) may be

a. 4i+3j b. 4i+8j c. 8i+4j d. -8i+4j

Because electric field at equi-potential surface is perpendicular to it. x&y coordinates must be of opposite sign. So correct answer would be D. And it is correct.

Even though I picked right answer for this question. I'm unable to solve it. Can any one help?

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2 Answers 2

up vote 1 down vote accepted

Yes, your approach is correct. In a generalised situation, one draws a normal to the equipotential lines to get the direction of field.

There is no way to solve this for the value of $\vec{E}$, because you don't know the potential gradient at that point. If they had given "potential is $V$ at $y=2x$, and $V+\Delta V$ at $y=2x+1$, as well as specifying that the potential varies linearly (alternatively that the E field is constant at all points), then you could construct an expression for the potential at a general point $(x,y)$ (It comes out to be $\phi (x,y)=V+(y-2x)\Delta V$), you can use the equation $\vec{E}=-\vec{\nabla}\phi$ (The upside down triangle is a shorthand, $\nabla\phi=\frac{\partial\phi}{\partial x}\hat{i} + \frac{\partial\phi}{\partial y}\hat{j} +\frac{\partial\phi}{\partial z}\hat{k} $)

But for this problem, you can find the direction of the electric field at any point on y=2x only (you are not sure if the rest of the equipotential curves are parallel lines).

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In general, you should construct a vector tangential to the equipotential line in the relevant point (in your case it's just a vector along the line) and check if it is orthogonal to the electric field strength vector.

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