Electrostatics · Physics · TS EAMCET

The electric flux due to an electric field $\mathbf{E}=(8 \hat{\mathbf{i}}+13 \hat{\mathbf{j}}) \mathrm{NC}^{-1}$ through an area $3 \mathrm{~m}^{2}$ lying in the $X Z$-plane is

A proton and an $\alpha$-particle are both accelerated from rest in a uniform electric field. The ratio of work done by the electric field on the proton and the $\alpha$-particle in a given time is

Two point charges $-10 \mu \mathrm{C}$ and $-5 \mu \mathrm{C}$ are situated on $X$-axis at $x=0$ and $x=\sqrt{2} \mathrm{~m}$. The point along the $X$-axis, where the electric field becomes zero is

Two point charges of magnitudes $-8 \mu \mathrm{C}$ and $+32 \mu \mathrm{C}$ are separated by a distance of 15 cm in air. The position of the point from $-8 \mu \mathrm{C}$ charge at which the resultant electric field becomes zero is

Two positive point charges are separated by a distance of 4 m in air. If the sum of the two charges is $36 \mu \mathrm{C}$ and the electrostatic force between them is 0.18 N , then the bigger charge is

A clock dial has point charges $-q,-2 q,-3 q, \ldots \ldots \ldots,-12 q$ at the positions of the corresponding numbers on the dial respectively. The time at which the hour's hand points the direction of the net electric field at the centre of the dial is (Assume clock hands do not influence the net electric field)

The electric potential at a place is varying as $V=\frac{1}{2}\left(y^2-4 x\right) \mathrm{V}$. Then, the electric field at $x=1 \mathrm{~m}$ and $y=1 \mathrm{~m}$ is