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Question
Solve numerical example.
A potential difference of 5000 volts is applied between two parallel plates 5cm apart a small oil drop having a charge of 9.6 ×10-19 C falls between the plates. Find
- electric field intensity between the plates and
- the force on the oil drop.
Solution
Given: V = 5000 volts, d = 5 cm = 5 × 10-2 m, q = 9.6 × 10-19 C
To find: a. Electric field intensity (E)
b. Force (F)
Formulae: i. E = `"V"/"d"`
ii. E = `"F"/"q"`
Calculation: From formula (i),
E = `5000/(5xx10^-2)` = 105 N/C
From formula (ii)
F = E × q
= 105 × 9.6 × 10-19
= 9.6 × 10-14 N
- Electric field intensity between the plates is 105 N/C
- Force on the oil drop is 9.6 × 10-14 N
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