Two parallel plates at a distance of 8.0 x 10-3m apart are maintained at a p.d of 600 volts with the negative plate earthed. What is the electric field strength?
- 4.8 Vm-1
- 75.0Vm-1
- 4800.0Vm-1
-
75000.0Vm-1
The correct answer is: D
Explanation
E = \(\frac{voltage}{distance}\)= \(\frac{600}{8 \times 10^2}\)
= 75000.0Vm-1