Explanation
When switch S is pressed, +q charge induced on outer side of right plate of A flows to left plate of B. It induces – q charge on inner side of right plate of B and + q charge on outer side of same plate, which flows to earth. ∴ Charge on B = +q – q = zero
Electric field inside a cavity is zero i.e. EA = 0. Field inside a conductor is zero i.e. EB = 0 Outside the conductor, field is not zero i.e.EC ≠ 0.
As the insulator plate is passed between the plates of the capacitor, its capacity increases first and then decreases as the plate slips out. As a result, positive charge on plate A increases first and then decreases. Hence current in the outer circuit flows first from B to A and then from A to B
The same force will act on both bodies although their directions will be different.
Torque about Q of charge –q is zero, so angular momentum charge –q is constant but distance between charges is changing, so force is changing, so speed and velocity are changing.
We put a Unit positive charge at 0. Resultant force due to the charge placed at A and C is zero and resultant charge due to B and D is towards D along the diagonal BD.
Because in case of metallic sphere either solid or hollow, the charge will reside on the surface of the sphere. Since both spheres have same surface area, so they can hold equal maximum charge.
Electric field inside a conductor is zero.
Change on glass rod is positive, so change on gold leaves will also be positive. Due to X-rays, more electrons from leaves will he emitted, so leaves becomes more positive and diverge further.
Please disable the adBlock and continue. Thank you.