Question

Two sphere of charge +10 and +40c are placed .12m a part.Find the position of the point between them ,when the electric field intensity is zero.


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Answer 1

Let a charge of +10C placed at point A and of charge +40C paced at point B and a point between them where third charged placed be Q.

And the distance between both the charges is 0.12 m i.e. distance between A and B is 0.12 m.

Now, assume that distance x from charge +10C.

→ F = (k q1 q2)/r²

F1 = [k (10)(Q)]/x² .....(eq 1)

F2 = [k (40)(Q)]/(0.12 - x)² ......(eq 2)

→ F1 = F2

→ [k (10)(Q)]/x² = [k (40)(Q)]/(0.12 - x)²

k and Q cancel throughout

→ 10/x² = 40/(0.12 - x)²

→ 10(0.12 - x)² = 40x²

→ 10(0.0144 + x² - 0.24x) = 40x²

→ 1(0.0144 + x² - 0.24x) = 4x²

→ 0.0144 + x² - 0.24x = 4x²

→ 3x² + 0.24x - 0.0144 = 0

→ 3(x² + 0.08x - 0.0048) = 0

→ x² + 0.08x - 0.0048 = 0

Using quadratic formula

D = b² - 4ac

→ (0.08)² - 4(1)(-0.0048)

→ 0.64 + 4(0.0048)

→ 0.0064 + 0.0192 = 0.0256

x = (-b ± √D)/2a

→ (-0.08 ± √0.0256)/2

→ (-0.08 ± 0.16)/2

→ (-0.08 + 0.16)/2

[negative one is neglected]

→ 0.04

Therefore, x is 0.04 m away from a charge Q.

$\rule{200}2$

Coulomb's Law

According to it magnitude of force between two static point charges is

• directly proportional to product of magnitude of charges

i.e. F $\alpha$ q1 q2

• inversely proportional to square of distance between two charges

i.e. F $\alpha$ 1/r²

→ F $\alpha$ (q1 q2)/r²

→ F = (k q1 q2)/r²