CBSE Class 12 · 2005 · ElectrostaticsShort Answer

Three point charges of

+ 2 μ C

- 3 μ C

and

- 3 μ C

are kept at the vertices A, B and C respectively of an equilateral triangle of side 20 cm as shown in the figure. What should be the sign and magnitude of the charge to be placed at the mid-point (M) of side BC so that the charge at A remains in equilibrium?

Answer: A positive charge of magnitude $q \approx 3.9 μ C$ placed at M.

Solution

Each

- 3 μ C

charge at B and C attracts the

+ 2 μ C

at A.
Magnitude of each force:

F_{1} = \frac{k ( 2 \times 1 0 ^{- 6} ) ( 3 \times 1 0 ^{- 6} )}{( 0.20 ) ^{2}} = 1.35 N

The two forces (along AB and AC,

6 0^{\circ}

apart) have resultant:

F = F_{1}^{2} + F_{1}^{2} + 2 F_{1}^{2} cos 6 0^{\circ} = 1.35 3 = 2.34 N

directed from A towards the mid-point M.
For equilibrium the charge at M must repel A (hence positive) with an equal force along MA.
With

A M = 0.10 3 m

\frac{k ( 2 \times 1 0 ^{- 6} ) q}{( 0.10 3 ) ^{2}} = 2.34

q = \frac{2.34 \times 0.03}{18 \times 1 0 ^{3}}

q \approx 3.9 \times 1 0^{- 6} C (positive, at M)

CBSE Class 12 Board · 2005 · Subjective

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CBSE Class 12 · 2005 · ElectrostaticsShort Answer

Solution

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