Question

Brainlist ❤️ Question ❓

Q 1.1) What is the force between two small charged spheres having charges of 2 × 10-⁷C and 3 × 10-⁷C placed 30 cm apart in the air?


Q 1.2) The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in the air is 0.2 N.

(a) What is the distance between the two spheres?

(b) What is the force on the second sphere due to the first?
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Answer 1

$\huge\bf\underline{\underline\red{Answer}}$

Q 1.1)

$\rightarrow$Repulsive force of magnitude 6 × 10 - 3 N

$\rightarrow$Charge on the first sphere, q1 = 2 × 10 - 7 C

$\rightarrow$Charge on the second sphere, q2 = 3 × 10 - 7 C

$\rightarrow$Distance between the spheres, r = 30 cm = 0.3 m

Electrostatic force between the spheres is given by the relation,

$\ \sf \: F = \frac{q1 q2 } {4 π ε0 r²}$

Where, ∈0 = Permittivity of free space

$\sf \frac{ 1} {4π ε 0} = 9 × 10⁹ N m² C -²$

$\sf \: F = \frac {9 × 10⁹ × 2 × 10 -⁷ × 3 × 10 -⁷ }{ ( 0.3 )²} = 6 × 10 -³ N$

Hence,

Force between the two small charged spheres is 6 × 10 - 3 N.

The charges are of same nature.

$\therefore$ force between them will be repulsive.

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Q 1.2)

A)

$\hookrightarrow$ Let charge on small sphere is denoted

by q1

Charge on large sphere is denoted by q2

force is denoted by F

Given :

q1 = 0.44C = 0.4 x 10-C

q20 8uC = 0.8 x 10-C

F 0.2 N

by using Coulombs law,

$\sf \: F= \frac{K \: q1q2}{r²}$

$\sf \frac{0.2 9 \: x \: 10⁹ \: x \: 0.4 \: x \: 10 -⁶ \: \: x \: 0.8 \: x \: 10 -⁶ }{r²}$

$\sf \: r {}^{2} = \frac{(9x 0.32) x 10 -³}{0.2}$

r² = 14.4 x 10 -³

r² = 144 x 10 -⁴

by taking square root both sides,

r 12 x 10-² m 12cm

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B )

$\hookrightarrow$ According to Newton's third law,

| F12 | = | F21 |

so,

F12 = F21 = 0.2N

$\huge\rm\fbox\blue{Thanks}$