Question

figure shows a set of equipotential surfaces. the magnitude and direction of electric field that exists in the regiin is

Answer 1

we know, electric field is the negative of potential gradient. e.g., $E=-\frac{\Delta V}{\Delta X}$

here negative sign indicates that electric field intensity will be increase in the direction of decreasing potential.


here, ∆X be the separation between two successive equipotential surfaces and ∆V be the difference of potential between two successive surfaces.

∆V = 30 - 40 = 20 - 30 = 10 - 20 = -10V
and sin45° = ∆X/1
or, ∆X = sin45°

so, electric field = -(-10)/sin45° = 10√2 V/m
therefore Magnitude of electric field is 10√2 V/m and direction of electric field 45° with x axis

Answer 2

Answer:

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