JEE Mains Model Paper 22

The angle between potential Edl = dv
Change in potential = 10V = dV
As E = ⊥ r dV (As potential surface)

So, Edl = dv ⇒ Edlcos(90° ≠ 30°) = -dv
⇒ E (10 × 10−2) cos 120° = -dv
⇒ E =

−dv

10×10−2cos120°

= −

10

10−1×(−1∕2)

= 200 V/m making as angle 120°
with x axis