Question

1.)The electric field between two square metal plates is 470N/C . The plates are 1.8m on a side and are separated by 4.0cm .What is the charge on each plate (assume equal and opposite)? Neglect edge effects.

2.)The field just outside a 3.70

cm-radius metal ball is 2.75

Answer 1

I have solved this question earlier with different figures. Please workout using yours figures. If you need any further help just PM me. If I have helped you please rate me 5 stars first (before you rate anyone else)

The electric field between two square metal plates is 160 N/C. The plates are 1.9 m on a side and are separated by 1.2 cm. What is the charge on each plate? Neglect edge effects.

Q=___ nC   

Answer

Electric Field,E = QA/?

Thus

Q = E?/A = (160*8.85*10^-12)/(1.9*1.9)

Q = 3.92*10^-9C = 3.92 nC

Answer 2

1)
E= 470 N/C
eo=8.85*10^12

E= Q / eo * area
Q = 470*8.85*10^-12*(1.8)^2 =13.476 *10^-9 C

2)
for the ball electric field on its surface will be
E = 9x10^9 x q / R^2 , where R is its radius.then,

charge on the ball q = E x R^2 / 9x10^9
= 275 x 3.75x3.75x10^-4 / 9x10^9

= 42.9 x10^-12 C. ( nearly)

Answer 3

I have solved this question earlier with different figures. Please workout using yours figures. If you need any further help just PM me. If I have helped you please rate me 5 stars first (before you rate anyone else)

The field just outside a 4.01 cm-radius metal ball is 7.26