electric field Questions

A charged oil drop is suspended in a uniform field of $3\times 10^4\text{V/m}$ so that it neither falls nor rises. The charge on the drop will be: (Take mass of the drop $=9.9\times 10^{-15}\text{kg}$ and $g=10{\text{m/s}}^2$)

Charges $+q,+q,-q,-q$ are placed at the vertices $A,B,C,D$ of a square respectively. Let $\vec{E}$ be the electric field and $V$ the potential at the centre. If the charges on $A$ and $B$ are interchanged with those on $D$ and $C$ respectively, then:

The potential at a point $x$ (measured in $\mu \text{m}$) due to some charges situated on the $x$ -axis is given by $V(x)=20/(x^2-4)$ Volts. The electric field $E$ at $x=4\mu \text{m}$ is given by:

Two point charges $+8q$ and $-2q$ are located at $x=0$ and $x=L$ respectively. The location of a point on the x-axis at which the net electric field due to these two point charges is zero is:

A charged ball $B$ hangs from a silk thread $S$ which makes an angle $\theta$ with a large charged conducting sheet $P$, as shown in the figure. The surface charge density $\sigma$ of the sheet is proportional to:

A particle of charge $-16\times 10^{-18}\text{C}$ moving with velocity $10\text{m/s}$ along the $x$ -axis enters a region where a magnetic field of induction $B$ is along the $y$ -axis, and an electric field of magnitude $10^4\text{V/m}$ is along the negative $z$ -axis. If the charged particle continues moving along the $x$ -axis, the magnitude of $B$ is:

A thin spherical shell of radius $R$ has charge $Q$ spread uniformly. Which graph represents the electric field $E(r)$ for $0\le r<\infty$?

Let $\rho (r)=\frac{Qr}{\pi R^4}$ be the charge density distribution for a solid sphere of radius $R$ and total charge $Q$. For a point '$p$ ' inside the sphere at distance $r_1$ from the centre of the sphere, the magnitude of electric field is:

In a uniformly charged sphere of total charge $Q$ and radius $R$, the electric field $E$ is plotted as a function of distance $r$ from the centre. The graph which would correspond to the above will be: