Home
Class 13
PHYSICS
Two fixed charges "A" and "B" of 5 mu C ...

Two fixed charges "A" and "B" of `5 mu C` each are separated by a distance of "6m" .C is the mid point of the line joining "A" and "B" .A charge Q of `-5 mu C` is shot perpendicular to the line joining "A" and "B" through "C" with a kinetic energy of "0.06" J. The charge 'Q' comes to rest at a point "D" .The distance "CD" is (a) 3m (b) `sqrt(3)m` (c) `3sqrt(3)m`(d) `4m`

Promotional Banner

Similar Questions

Explore conceptually related problems

Two point charges of 5 mu C and 20 mu C are separated by a distance of 2m. Find the point on the line joining them at which electric field intensity is zero.

Two point charges 4 mu C and -2 mu C are separated by a distance of 1m in air. Calculate at what point on the line joining the two charges is the electric potential zero ?

Two point charges +5mu C and -2mu C are kept at a distance of 1m in free space. The distance between the two zero potential points on the line joining the charges is

Two positive charges of 1mu C and 2 mu C are placed 1 metre apart. The value of electric field in N/C at the middle point of the line joining the charge will be :

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

The force between two charges 9 mu C and 1mu c separated by a distance 3 m in space is..........

Two tiny spheres carrying charges 1.8 muC and 2.8 mu C are located at 40 cm apart. The potential at the mid-point of the line joining the two charges is

Two point charges of 1muC and -1muC are separated by a distance of 100 Å. A point P is at a distance of 10 cm from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at P will be

Show that the line joining the point A(1,-1,2) and B(3, 4, -2) is perpendicular to the line joining the points C(0,3,2) and D(3,5,6).

Two charges 5 mu C and 4 mu C are separated by a distance 20cm in air. Work to be done to decrease the distance to 10cm is (x)/(10)J .Where x is........