The potential in an electric field varies as \(V=\left(x^2-y^2\right)\). The electric lines of the force in…
The potential in an electric field varies as \(V=\left(x^2-y^2\right)\). The electric lines of the force in \(X-Y\) plane are
Solution
The potential in an electric field varies as,
\(V=\left(x^2-y^2\right)\)
\(\therefore\) Electric field,
\(\begin{aligned}
E & =-\left[\frac{d v}{d x} \hat{\mathbf{i}}+\frac{d v}{d y} \hat{\mathbf{j}}\right] \quad[\because E=\Delta V] \\
& =-\left[\frac{d}{d x}\left(x^2-y^2\right) \hat{\mathbf{i}}+\frac{d}{d y}\left(x^2-y^2\right) \hat{\mathbf{j}}\right] \\
E & =-[2 x \hat{\mathbf{i}}-2 y \hat{\mathbf{j}}] \Rightarrow E=-2 x \hat{\mathbf{i}}+2 y \hat{\mathbf{j}}
\end{aligned}\)
Expression of the electric field is linear equation in two variables, i.e. straight lines in \(X-Y\) plane and slope having \(45^{\circ}, 35^{\circ}\) etc.
Hence, the option (c), represents correct graph.
