Question

A positive charge is placed between the plates of a parallel plate capacitor and released from rest at Point B, as shown in the figure. In what direction does the charge move? O The charge moves toward Point B O The charge moves toward Point C. O The charge remains at rest. In the previous question, the work done by the electrostatic force in moving the positive charge from Point B to Point C (WBc) would be equal to which of the following? O The electrostatic potential energy at Point B (EPEg) The electrostatic potential energy at Point C (EPEc) The difference in the electrostatic potential energy between Points B and C(EPEB EPEc). O The sum of the electrostatic potential energy between Points B and C (EPEg + EPEC).

Answer 1

The electric field is in the direction of positive to negative charge and a positive charge moves in the direction of electric field.

The electric field is in the direction of positive to negative charge and a positive charge moves in the direction of electric field.� Due to opposite charges on the plates, an electric field is produced directed from the positive plate to the negative plate as shown in the above figure� implies A positive charge placed in the electric field will move in the direction of the electric field. (that is downwards point c)� Hence,� The charge moves towards point c� The electric field produces on potential gradient, given by� E = -dv/dr� And the charge will move due to the potential difference.� Hence the negative sign shows that, the motion in equal to the� The difference in the potential energy between points B and C.� i.e (E_p E_D - F_p E_c)