Question

Three point charges of +2 μC, -3 μC and -3 μC are kept at the vertices A, B and C respectively of an equilateral triangle of side 20 cm as shown in 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?

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Answer 1

Three point charges of +2 μC, -3 μC and -3 μC are kept at the vertices A, B and C respectively of an equilateral triangle of side 20 cm as shown in 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 2

Given:

charge +2μC is placed at vertice A, charge -3μC at point B and charge -3μC is placed at C of a equilateral triangle of side 20cm.

To find:

Sign and magnitude of the charge at midpoint M at BC.

Solution:

Let charge q₁ be placed at point A. q₁ = +2μC

     charge q₂ be placed at point B. q₂ = -3μC

     charge q₃ be placed at point C. q₃ = -3μC

Formula used is

$\vec{F_{AB}}= \frac{q_{1} q_{2}}{4\pi \epsilon _{0}(AB)^{2}}$

AB = distance of side AB=20cm=0.02m

putting all the values we get ,

$\vec{F_{AB}}= \frac{9 \times 10^9 \times (2 \times 10^-^6) \times (3 \times 10^-^6)}{(0.02)^2}$

$\vec{F_{AB}} = 1.35N$

Now, find

$\vec{F_{AC}}= \frac{q_{1} q_{2}}{4\pi \epsilon _{0}(AC)^{2}}$

$\vec{F_{AC}}= \frac{9 \times 10^9 \times (2 \times 10^-^6) \times (3 \times 10^-^6)}{(0.02)^{2}}$

$\vec{F_{AC}} = 1.35N$

M  is the mid point of BC. so,

$\vec{F_{AM}}= \frac{q_{1} q}{4\pi \epsilon _{0}(AM)^{2}}$

$\vec{F_{AM}}= \frac{9 \times 10^9 \times (2 \times 10^-^6) \times q}{(\sqrt{3} \times 10^-^3 )^{2}}$

$\vec{F_{AM}} = 6 \times 10^5 (q)$

For equilibrium charge at A, then

$F_{AB}cos30^{\circ} + F_{AC}cos 30^{\circ} = F_{AM}$

$1.35 \times \frac{\sqrt{3} }{2} + 1.35 \times \frac{\sqrt{3} }{2} = 6 \times 10^5 (q)$

therefore,

$q = \frac{1.35\sqrt{3} }{6 \times 10^5}$

q = 0.3897 × 10⁻⁵ C

So, charge has magnitude of 0.3897 × 10⁻⁵C.