✒✒✒Two small spheres A and B of charges 18 esu and 8 esu respectively,are placed at a distance 20 cm apart in air. At which point, theelectric field intensity will be zero ???✒✒✒ ___Ok sis ___ report my this question and fast question please About the author Reagan
Answer: Initially 2nC and −8nC are separated by a distance d. Electrostatic force acting between them initially F= d 2 kq 1 q 2 ∴ F= d 2 (2)(8)k = d 2 16k Now the two charges are allowed to touch each other, they share equal charges among themselves. Total charge of the system Q=2−8=−6nC Thus new charge on each sphere q 1 ′ =q 2 ′ =−3nC Let the new distance between them be d ′ . Thus force acting between them F ′ = (d ′ ) 2 kq 1 ′ q 2 ′ ∴ F ′ = (d ′ ) 2 k(−3)(−3) = (d ′ ) 2 9k But F ′ =F ∴ (d ′ ) 2 9k = d 2 16k Or (d ′ ) 2 = 16 9 d 2 ⟹ d ′ = 4 3d Reply
Answer:
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Answer:
Initially 2nC and −8nC are separated by a distance d.
Electrostatic force acting between them initially F=
d
2
kq
1
q
2
∴ F=
d
2
(2)(8)k
=
d
2
16k
Now the two charges are allowed to touch each other, they share equal charges among themselves.
Total charge of the system Q=2−8=−6nC
Thus new charge on each sphere q
1
′
=q
2
′
=−3nC
Let the new distance between them be d
′
.
Thus force acting between them F
′
=
(d
′
)
2
kq
1
′
q
2
′
∴ F
′
=
(d
′
)
2
k(−3)(−3)
=
(d
′
)
2
9k
But F
′
=F
∴
(d
′
)
2
9k
=
d
2
16k
Or (d
′
)
2
=
16
9
d
2
⟹ d
′
=
4
3d