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Column – I shows graphs of electric potential V versus x and y in a certain region for four situations. Column – II gives the angle which the electric field vector may make with positive x-direction.
Column – I
(V versus x) (V versus y)
Column – II
A)
P) $0 \leq θ < 45 °$
B)
Q) $45 ° \leq θ < 90 °$
C)
R) $90 ° \leq θ < 135 °$
D)
S) $135 ° \leq θ \leq 180 °$

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A – P; B – S; C – R; D – Q

A – Q; B – R; C – P; D – S

A – S; B – P; C – R; D – Q

A – S; B – P; C – Q; D – R

answer is C.

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Detailed Solution

The x component of electric field is given by the slope of V - x graph and y component is given by the slope of V - y graph. For example, in option (A) of column – I
$E_{x} = - \frac{\partial V}{\partial x} = - 1 a n d E_{y} = - \frac{\partial V}{\partial y} = \tan 30 ° = \frac{1}{\sqrt{3}}$
Now, it is easy to work out the direction of E.

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