Question

1) Calculate the magnitude of the electric field 2.12m from a point charge of 6.4mC (such as found on the terminal of a Van de Graaff).
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2) The electric field strength between two parallel conducting plates separated by 7.0cm is 8.5x10^4 V/m. The potential difference between the plates is 6kV. The with the lowest potential is taken to he zero volts. What is the potential 1.1cm from that plate?

Answer 1

1) given that

charge q = 6.4*10^-3 C

distance r = 2.12 m

from the relation

electric field due to a point charge E = kq/r^2

E = 9*10^9*6.4*10^-3/2.12^2 = 12.82*10^6 N/C

2) given

plates separation d = 0.07 m

electric field E = 8.5*10^4 V/m

potential difference = $\Delta$ V = 6*10^3 V

from the relation

change in potential = electric field*distance

v-0 = 8.5*10^4*1.1*10^-2

v = 0.935 kV