Embibe Experts Solutions for Exercise 3: EXERCISE 2.3

A rod is bent into a semi-circular arc of radius $R.$ The rod has a uniform linear charge distribution $\lambda .$ The potential at the centre of the arc, point $P$ is

An electric field exists in space as $E=2\hat{i}+3\hat{j}+0\hat{k}.$ Find the potential difference $\left(V_A-V_B\right)$ between points $A(2, 2, 0)$ and $B(1, 1, 0).$

The electric potential in space is given by $V(x, y, z)=\left(2xy+y^{2}z+x^{2}yz\right)$ volt, where $x, y$ and $z$ are in metre. Then $y$ component of electric field at the point $(3, -2, 1)$ will be

The electrostatic potential difference between points $A$ and $B, V_A-V_B,$ which are distances $r_A=2.0 \mathrm{m}$ and $r_B=1.0 \mathrm{m}$ from an infinitely long thin wire $\lambda =1.0\mu \mathrm{C}/\mathrm{m}$ is

In the above situation (i.e., Q. 27), if an electron is released from rest at point $A$ it's speed at point $B$ will be

A uniform electric field having a magnitude $E_0$, directed along positive x-axis exists. If the electric potential $V$ is zero at $x=0$, then its value at $x=+x$ will be

The electric potential$V$at any point $\left(X, Y, Z\right)$ in space is given by $V=4x^2\text{ V}$. the electric field at$\left(1, 0, 2\right)$ m in ${\text{Vm}}^{-1}$ is

In an electric field, potential at a point with position vector $\overrightarrow{r}$ is given as $v=\frac{343}{\left|\overrightarrow{r}\right|}$. The electric field at $\overrightarrow{r}=3\hat{i}+2\hat{j}+6\hat{k}$ is