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Ymorist [56]
3 years ago
13

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?
Physics
1 answer:
BlackZzzverrR [31]3 years ago
5 0

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

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For a concave mirror, for the sign convention f is considered to be positive. So we can solve the equation for d_i by using the numbers given in the text of the problem:
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