Answer:

Given:

A particle A of mas of 3mg is balanced by keeping ANOTHER object of B of charge +30nC below it at a depth of 5cm.

To find:

Charge on A in ORDER to balance the object against Gravity.

Calculation:

Object A will be balanced against gravity only when the gravitational force acting on it is equal to the electrostatic force exerted by object B.

\therefore \: F_{g} = F_{E}∴F

g

=F

E

= > mg = \dfrac{k(q1)(q2)}{ {r}^{2} }=>mg=

r

2

k(q1)(q2)

= > 3 \times {10}^{ - 6} \times 10 = \dfrac{k(q1)(30 \times {10}^{ - 9} )}{ {(0.05)}^{2} }=>3×10

−6

×10=

(0.05)

2

k(q1)(30×10

−9

)

= > 3 \times {10}^{ - 5} = \dfrac{k(q1)(30 \times {10}^{ - 9} )}{ {(0.05)}^{2} }=>3×10

−5

=

(0.05)

2

k(q1)(30×10

−9

)

= > 3 \times {10}^{ - 5} = \dfrac{k(q1)(30 \times {10}^{ - 9} )}{ 25 \times {10}^{ - 4} }=>3×10

−5

=

25×10

−4

k(q1)(30×10

−9

)

= > 3 \times {10}^{ - 5} = \dfrac{k(q1)(6 \times {10}^{ - 9} )}{ 5 \times {10}^{ - 4} }=>3×10

−5

=

5×10

−4

k(q1)(6×10

−9

)

= > {10}^{ - 5} = \dfrac{k(q1)(2 \times {10}^{ - 9} )}{ 5 \times {10}^{ - 4} }=>10

−5

=

5×10

−4

k(q1)(2×10

−9

)

= > {10}^{ - 9} = \dfrac{k(q1)(2 \times {10}^{ - 9} )}{ 5 }=>10

−9

=

5

k(q1)(2×10

−9

)

= > 1 = \dfrac{k(q1)(2 )}{ 5 }=>1=

5

k(q1)(2)

= > 1 = \dfrac{9 \times {10}^{9} (q1)(2 )}{ 5 }=>1=

5

9×10

9

(q1)(2)

= > \: q1 = \dfrac{5 \times {10}^{ - 9} }{18}=>q1=

18

5×10

−9

= > \: q1 = 0.27 \times {10}^{ - 9}=>q1=0.27×10

−9

= > \: q1 = 0.27 \: \: nC=>q1=0.27nC

So, final answer:

\boxed{ \SF{ \: q1 = 0.27 \: \: nC }}

q1=0.27nC