Positive and negative point charges of equal magnitude are kept at $\left(\mathrm{0,0},\frac{a}{2}\right)$ and $\left(\mathrm{0,0},\frac{-a}{2}\right)$ , respectively. The work done by the electric field when another positive point charge is moved from $\left(-a,\mathrm{ }0,\mathrm{ }0\right)$ to $\left(0,\mathrm{ }a,\mathrm{ }0\right)$ is 

As per this diagram a point charge: $+q$ is placed at the origin $O.$ Work done in taking another point path charge $-Q$ from the point $A$ [co-ordinates $\left(o, a\right)$] another point $B$ [co-ordinates $\left(a, o\right)$] along the straight path $AB$ is:

Two charges $q_1$ $q_2$ are placed $30 \mathrm{cm}$ apart, as shown in the figure. A third charge $q_3$ is moved along the arc of a circle of radius $40 \mathrm{cm}$ from $C$ to $D.$ The change in the potential energy of the system is $\frac{q_3}{4\pi {\epsilon }_0}k,$ where $k$ is:

Charges $+q$ and $-q$ are placed at points $A$ and $B$ respectively which are a distance $2L$ apart, $C$ is the mid-point between $A$ and $B.$ The work done in moving a charge $+Q$ along the semicircle $CRD$ is

A charged particle $\text{'}q\text{'}$ is shot towards another charged particle $\text{'}Q\text{'},$ which is fixed, with a speed $\text{'}v\text{'}.$ It approaches $\text{'}Q\text{'}$ up to a closest distance $r$ and then returns. If $q$ were given a speed $\text{'}2v\text{'},$ the closest distance of approach would be:

Two insulating plates are both uniformly charged in such a way that the potential difference between them is $V_2-V_1=20 \mathrm{V}.$ (i.e. plate $2$ is at a higher potential). The plates are separated by $d=0.1 \mathrm{m}$ and can be treated as infinitely large. An electron is released from rest on the inner surface of plate $1.$ What is its speed when it hits plate $2?$ $\left(e=1.6\times {10}^{-19}\mathrm{C}, m_e=9.11\times {10}^{-31} \mathrm{kg}\right)$