Question

A.) Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 2.5×10⁹ electrons from one disk to the other causes the electric field strength between them to be 1.0×10⁵ N/C .

What are the diameters of the disks? Express your answer to two significant figures and include the appropriate units.

B.) At a distance r from a point charge, the electric potential is 4600 V and the magnitude of the electric field is 2.0×10⁵ V/m .

- What is the distance r? Express your answer to two significant figures and include the appropriate units.

- What is the electric potential at distance r / 2 from the charge? Express your answer to two significant figures and include the appropriate units.

- What is the magnitude of the electric field at distance r / 2 from the charge? Express your answer to two significant figures and include the appropriate units.

Answer 1

here,

QUEs 1:

no of electrons, n = 2.5*10^9

Charge due to these electrons, Q = 1.6*10^-19 * 2.5*10^9
Charge due to these electrons, Q = 4*10^-10 C

Electric field, E = 10^5 N/C

Since, Charge in parallel plate capacitor is given as:

Q = eo * A * E

Area, A = Q/eo*E

pi * (d^2/4) = Q/eo*E

diamater, d = sqrt((Q * 4)/(eo * E * pi))
diamater, d = sqrt((4*10^-10 * 4)/(8.85*10^-12 * 10^5 * pi))
diamater, d = 0.024 m

QUES 2:

Electric field, E = 2*10^5 V/m

Electric potential, V = 4500 V

Electric potential du to electric field at point is given as:
v = r * E

distance to point, r = V/E = 4500 / (2*10^5) = 0.022 m

Electric potential, V = E * r = 2*10^5 * 0.022/2 = 2200 V

Electric field, E = V/r = 4500/(0.022/2) = 410000.000 V/m