Two charge `Q_(1)=18muC` and `Q_(2)=-2 mu C` are separated by a distance R, and `Q_(1)` is on the left off `Q_(2)`. The distance of the point where the net electric field is zero is

Two charge Q_(1)=18muC and Q_(2)=-2 mu C are separted by a distance R, and Q_(1) is on the left off Q_(2) . The distance of the point where the net electric field is zero is

Two positive point charges q_1=16muC and q_2=4muC are separated in vacuum by a distance of 3.0m . Find the point on the line between the charges where the net electric field is zero.-

Two point charges Q and - Q //4 are separated by a distance X . Then .

The two point charges 4muC and 1muC are separated by a distance of 2 m in air. Find the point on the line joining the charges at which net electric field of the system is Zero.

Two point charges, q_(1)=10xx10^(-8)C and q_(2)=-2xx10-8C are separated by a distance of 60 cm in air. (i) Find at what distance from the 1st charge, q_(1) , would the electric potential be zero. (ii) Also calculate the electrostatic potential energy of the system.

Two point charge q_(1)=2muC and q_(2)=1muC are placed at distance b=1 and a=2cm from the origin on the y and x axes as shown in figure .The electric field vector at point (a),(b) will subtnd on angle theta with the "x-axis" given by

Two point charges q_(1) = +2 C and q_(2) = - 1C are separated by a distance d . The position on the line joining the two charges where a third charge q = + 1 C will be in equilibrium is at a distance