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Home/Practice questions/Physics
PhysicsElectric Charges and Fields

Two point charges, +9e and +4e are placed in free space at a distance…

Two point charges, +9e and +4e are placed in free space at a distance of r from each other. Another charge Q is placed at a point on the line joining these two charges such that the system of charges stays at equilibrium. Consider the following statements about charge Q at equilibrium and identify the correct statement only:

(A) Magnitude of charge Q = 36e/25

(B) Distance of Q from +9e is 3r/5

(C) Distance of Q from +4e is 3r/5

(D) Charge Q is negative.

Choose the correct answer from the options given below:

Answer & explanation+

Correct answer: option 1

The correct answer is Option (1) → (A), (B) and (D) only

For the whole system to be in equilibrium, the net force on each charge must be zero. Let Q lie between the charges at distance x from +9e. The forces on Q balance when

    \( \displaystyle \frac{k(9e)Q}{x^{2}} = \frac{k(4e)Q}{(r - x)^{2}} \)

    \( \displaystyle \frac{3}{x} = \frac{2}{r - x} \Rightarrow x = \frac{3r}{5} \)

+9e repels +4e, so Q must pull it back: Q is negative. For zero net force on +4e:

    \( \displaystyle \frac{k(9e)(4e)}{r^{2}} = \frac{k|Q|(4e)}{(2r/5)^{2}} \)

    \( \displaystyle |Q| = 9e \times \frac{4}{25} = \frac{36e}{25} \)

(A) Magnitude of charge Q = 36e/25 — Correct. Shown above.

(B) Distance of Q from +9e is 3r/5 — Correct. \( x = 3r/5 \).

(C) Distance of Q from +4e is 3r/5 — Incorrect. It is \( r - 3r/5 = 2r/5 \).

(D) Charge Q is negative. — Correct. A positive Q would push both end charges outward.

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