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. What
should be the charge on A in order to balance it
against gravity?
Answers
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