Question

Three point charges 2mc,-3mc and -3mc are placed on the vertices of the equilateral triangle of the side 20cm.what should be the magnitude and sign of the charges placed at mid point of two negative charge so that 2mc is in equilibrium

Answer 1

the sign is positive and magnitude is 3mc

Answer 2

The magnitude and sign of the charges placed at a midpoint is 2.25$\sqrt{3}$ mc

Explanation:

Given:

Three point charges are 2mc, -3mc -3mc are vertices of equilateral triangle.

To find: The magnitude and sign of the charges of the equilateral triangle

Solution

Let the charge at M be $q_{m}$

The forces acting on charge +$q_{A} = -2 mc$

$F_{AB} , F_{AC},} F_{AM}$ are shown in the diagram

$F_{AB} = \frac{1}{4\pi E_{o} } \frac{q_{A} q_{B} }{r^{2}_{AB} }$   along AB

= $9$ × $10^{9} [ \frac{(2 *10)(3*10^{-6} }{(0.20)^{2} } ]$     along AB

where AB = 1.35 N

$F_{AC} = \frac{1}{4\pi E_{o} } \frac{q_{A} q_{C} }{r^{2}_{AC} }$

= 9 × $10^{9} [ \frac{(2 *10^{-6} )(3*10^{-6} }{(0.20)^{2} } ]$

Where AC = 1.35 N along AC

$F_{AM} = \frac{1}{4\pi E_{o} } \frac{q_{A} q_{M} }{r^{2}_{AM} }$

= 9 ×$10^{9} [ \frac{(2 *10^{-6} )q_{m} }{(\sqrt{3 }*10^{-1} )^{2} } ]$

= 6 ×$10^{5}q_{m}$ N  along MA

For the equilibrium charge q ,the resultant $F_{AB}$  and $F_{AC}$ must be equal and opposite to $F_{AM}$

Therefore

$F_{AB }cos 30 + F_{AC} cos 30 = F_{AM}$

⇒ 1.35 × $\frac{\sqrt{3} }{2} + 1.35\frac{\sqrt{3} }{2} = 6 * 10^{5} q_{m}$

$q_{m} = \frac{1.35\sqrt{3} }{6*10^{5} }$

= $0.225\sqrt{3} 810^{-5} c$

= 2.25$\sqrt{3}$mc

Final answer:

The magnitude and sign of the charges placed at a midpoint is 2.25$\sqrt{3}$ mc

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