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Question

what potential different is appliced to two metal plates .5m apart if the electric field between them is 2.5 times 10 to the third n/c?

Answer 1

Potential differences is voltage. Voltage means how much work can be done on moving a unit charge (or per charge).

Work = Fd
Electric field (E) is a force on a unit charge (q)
E=F/q
W/q=(F/q) d
we have

PD=V=W/q=Ed

V= 2.5E+3 x .5m=1.250 E+3 Volts