Question

A parallel-plate capacitor is formed from two 2.4 cm -diameter electrodes spaced 1.4 mm apart. The electric field strength inside the capacitor is 6.0×10^6 N/C. What is the charge (in nC) on each electrode?

Answer 1

Remember and apply for parallel plate caapcitor, these as formulas

1. Capacitance C = K eoA/d   

where K is dieelctric constant ( for air K = 1)

eo is permnittivity constant = 8.85 e-12

A is area = pi^2 ( for circular plates) and l* b for rectanular plates

and

d is the distance between the plates

-------------------------------------------------------------------------------------------------------
2.Charge Q = CV where V is potrntial difference

--------------------------------------------------------------------------------------------------------------

3. energy stored U = 0.5 QV or 0.5 CV^2 or 0.5 Q^2/C
-----------------------------------------------------------------------------------------------------------

4. electric field between the plates is E = Q/eo A ( also given by E = V/d)

so

Charge Q = E eo A

Q = 6 e 6 * 8.85 e -12 * 3.14 * 0.012* 0.012

Q = 24 nC --------------<<<<<<<<<<<<Answer