Question

Consider horizontal parallel plates with a fixed potential difference. The upper plate has a voltage difference of 30 V with the lower plate, and they are separated by 3 cm. You move the two plates carefully to a separation of 4 cm. What is the strength of the electric field between the plates?

Answer 1

relation between potential difference and electric field is given as

$E . d = \Delta V$

so here we know that

d = 3 cm

$\Delta V = 30 V$

$E \times 0.03 = 30$

$E = 1000 N/C$

So now when plates are separated to 4 cm distance carefully

the potential difference between them will change but the electric field between them will remain constant

So at distance of 4 cm also the electric field will be E = 1000 N/C