Please show work and explain Thank you A parallel plate capa

Please show work and explain. Thank you.

A parallel plate capacitor with radii 15.0 cm and separation distance 5.00 cm, is charged to Q = 25 nC. A test charge, q, having mass 2.50 g hangs on a string and makes and angle of 30.0Degree from the vertical, what is the charge q on this test charge?

Solution

the capacitance of the capacitor is,

    C = e₀A/d = 8.85*10^-12*pi*0.15*0.15/0.05 = 1.250505e-11 F

The potential difference is,

    V = Q/C = 25*10^-9 / 1.250505e-11 = 1.999e+3 V

The electric field is,

   E = V/d = 1.999e+3 / 0.05 = 3.998e+4 N/C

the charge is calculated as follows:

   Fcos30 = mg

   Fsin30 = qE

Hence,

     sin30/cos30 = qE/mg

hence, the charge q is,

   q = mgtan30/E = 2.5*10^-3*9.8*tan30/ 3.998e+4 = 3.54e-7 C = 0.354e-6 = 0.354 uC = 354 nC