The electric potential V at a certain point distant x metre from the origin is given by V(x)= (5x2+10x-9) volt. The electric field at x=1m will be
20 v/m
-10 v/m
10 v/m
-20v/m
Two charged balls attract each other with a certain force.If they are allowed to touch and then separated to the same distance appart the two balls will
Attract with smaller force
Attract with greater force.
Repel with smaller force
Repel with greater force.
Two conducting spheres of radii 5cm and 3cm are charged to the same potential. The ratio of their charges will be
3:5
5:3
9:25
25:9
Two spherical conductors B and c of equal radii and carrying equal charge on them repel each other with a force F when kept apart at some distance.A third spherical conductor having some redius as that of B but uncharged is brought in contact with B and then with C and removed away from both.The new force of repulsion between B and C
F/4
3F/4
F/8
3F/8
In a charged capacitor the energy is stored
In the filed between the plates
In the positive charges
In the negative charges
On the plates
A copper slab of thickness d/2 is introduced between the plates of a parallel plate capacitor of plate separation d. If the capacitance of the capacitor without copper slab is c and with copper slab is c', then c '/c is
√2
2
1
1/√2
A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be
1/4
1/2
For what purpose Gold leaf electroscope is used?
To find the electric field
To find the electric potential
To detect the charge on a body
To find the force between charges.
Two charges q1 and q2 are seperated by a distance d in air.A dielectric slab of thickness t of relative permittivity k is introduced between them.The force between the two charges is nearly given by
1/4πεo q1q2/(d-t)2
1/4πεo q1q2/(d-t+t/k)2
1/4πεo q1q2/(d-t+t√k)2
1/4πεo q1q2/(d+kt)2
