An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is
Smaller
5 times greater
Equal
10 times greater
A certain charge Q is divided into two parts q and Q-q. How the charge Q and q must be related so that when q and (Q-q) is placed at a certain distance apart experience maximum electrostatic repulsion?
Q = 2q
q = 3q
Q = 4q
Q= 4q + c
A charged particle 'q' is shot with speed v towards another fixed charged particle Q. It approaches Q upto the closest distance r and then returns. If q were given a speed 2v, the closest distance of approach would be
r
2r
r/2
r/4
A conducting sphere of radius R = 20 cm is given by a charge Q = 16μC. What is E at centre?
3.6×106 N/C
1.8×106 N/C
Zero
0.9×106 N/C
If an insulated non-conducting sphere of radius R has charge density ρ. The electric field at a distance r from the centre of sphere (r > R) will be
Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged, is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is
Three equal charges, each having a magnitude of 2.0 x 10-6 C, are placed at the three corners of a right angled triangle of sides 3 cm, 4 cm and 5 cm. The force (in magnitude) on the charge at the right angled corner is
50 N
26 N
29 N
45.9 N
Two charges of + 10 μC and +20 μC are separated by a distance 2 cm. The net potential (electric) due to the pair at the middle point of the line joining the two changes, is
27 MV
18 MV
20 MV
23 MV
The charges on two spheres are + 7 µC and -5C respectively. They experience a force F. If each of them is given an additional charge of -2 µC, then the new force of attraction will
F
2F
A.
F
Given:-
q1 = + 7gµC = +7 X 10-6 C
q2 = -5 μC = -5 × 10-6 C
New force F'= ?
We know that,
F =
F =
= N
F' =
= N
⇒ F' = F