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Question
A charge +q exerts a force of magnitude -0.2 N on another change -2q. If they are separated by 25.0 cm, determine the value of q.
Solution
Given: q1 = + q, q2 =−2q, F = – 0.2 N, r = 25 cm = 25 × 10−2 m
To find: Charge (q)
Formula: F = `1/(4πε_0)("q"_1"q"_2)/"r"^2`
Calculation: From formula,
– 0.2 = `(9xx10^9xx"q"xx(-2"q"))/((25xx10^-2)^2)`
∴ `(- 0.2)/(- 2) xx (25 xx 10^-2)^2/(9 xx 10^9) = "q"^2`
∴ `0.1 xx (25 xx 10^-2)^2/(9 xx 10^9) = "q"^2`
∴ `1/10 xx (25 xx 10^-2)^2/(9 xx 10^9) = "q"^2`
`∴ (25 xx 10^-2)^2/(9 xx 10^10) = "q"^2`
∴ `sqrt((25 xx 10^-2)^2/(9 xx 10^10))` = q
∴ `(25 xx 10^-2)/(3 xx 10^5)` = q
∴ q = 8.33 × 10−7 C
or q = 0.833 × 10−6 C
∴ q = 0.833 μC
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