The potential is varying with distance ( {\x}) as {\V}=\frac{1}{2} ( {\y}^2-4 {\x}) volt. The electric field at x=1 {\m} and y=1 {\m} is

Correct Answer is : 2^i−^jVm−1

2 \stackrel{\^}{ {\i}}+\stackrel{\^}{ {\j}} {\Vm}^{-1}

-2 \stackrel{\^}{i}+\stackrel{\^}{j} {\Vm}^{-1}

2 \stackrel{\^}{i}-\stackrel{\^}{j} {\Vm}^{-1}

-2 \stackrel{\^}{ {\i}}+2 \stackrel{\^}{ {\j}} {\Vm}^{-1}

AP EAPCET 17th May 2023 Shift 2 Solved Paper

Section: Physics

Question : 105 of 160

Marks: +1, -0

The potential is varying with distance (x) as V=‌

(y2−4x) volt. The electric field at x=1m and y=1m is

Go to Question:

Prev Question

Next Question