Question

Two small equal and unlike charges 3 X 10-8 are placed at A and B at

a distance of 6cm. Calculate the force on the charge +1 X10-8 placed at P,

where P is at 4cm on the perpendicular bisector of AB.​

Answer 1

Correct representation: Two small equal and unlike charges 3 × 10⁻⁸ are placed at A and B at a distance of 6 cm. Calculate the force on the charge +1 × 10⁻⁸ placed at P, where P is at 4 cm on the perpendicular bisector of AB.​

The force on the charge +1 × 10⁻⁸ placed at P is 1.296 × 10⁻³ N.

Given:

Two small equal and unlike charges = 3 × 10⁻⁸

Charge placed at P = +1 × 10⁻⁸

To Find:

The force on the charge +1 × 10⁻⁸ placed at P =?

Solution:

Unlike charges attract each other. It means they have opposite signs.

Let's take two equal and unlike charges q1 and q2 placed at A and B respectively.

q1 = 3 × 10⁻⁸ and q2 = -3 × 10⁻⁸

Let the charge at P be q.

q = 1 × 10⁻⁸

The distance between A and B is given as 6cm.

P is perpendicular to AB and meets AB at Q. It divides AB in half.

So, AP = PB = 3cm.

The distance between PQ is given as 4 cm.

Triangle PQA forms a right-angle triangle.

Using the Pythagorean theorem, we can find the AP.

AP² = AQ² + PQ²

AP² = 3² + 4²

AP² = 9 + 16

AP² = 25 = 5²

∴ AP = 5 cm or 5 × 10⁻² m

According to Coulomb's law, the magnitude of force F of two charges q1 and q2 having a distance r between them is given by

F = $\frac{1}{4\pi\epsilon o\ }$$\frac{q1.q2}{r^{2} }$

$\frac{1}{4\pi\epsilon o\ }$ = 9 × 10⁹ Nm²/C²

When two charges have the same sign, they repel each other. Charge q1 will repel charge q with force F.

F = $\frac{1}{4\pi\epsilon o\ }$$\frac{q1.q}{r^{2} }$

$F = \frac{9*10^9*3*10^{-8}*1*10^{-8} }{({5 * 10^{-2}})^2 } \\F = \frac{27*10^{-7}}{25 * 10^{-4} }$

F = 1.08 × 10⁻³ N along AP.

When two charges have opposite signs, they attract each other. Charge q2 will attract charge q with the same force F because AP = BP = 5cm.

So, the force here will be the same as the force along AP.

F = 1.08 × 10⁻³ N along BP.

To find the force on the charge +1 × 10⁻⁸ placed at P, we resolve the force into two components.

Net force = Fcosθ + Fcosθ = 2Fcosθ

∠PAQ of ΔPQA is θ

cosθ = base/hypotenuse

base = 3 cm and hypotenuse = 5 cm.

So, cosθ = 3/5

Resultant force = 2 × 1.08 × 10⁻³ × $\frac{3}{5}$

= 2.16 × 10⁻³ × $\frac{3}{5}$

= 1.296 × 10⁻³ N

Hence, the force on the charge +1 × 10⁻⁸ placed at P is 1.296 × 10⁻³ N.

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