Question : Two identical spheres when placed 3m apart repel each other with a force of 8 mN. When they are joined by a wire they repel each other with a force of 9 mN. Calculate their initial charges.
Doubt by Deepanshi
Solution :
Case I
q1 and q2 are the initial charges present on two identical spheres.
Case I
q1 and q2 are the initial charges present on two identical spheres.
r = 3m
F = 8 mN = 8×10-3 N
As per the coulomb's law
F= kq1q2/r²
Fr²/k = q1q2
q1q2 = Fr²/k
q1q2 = [(8×10-3)×(3)²]/9×109
q1q2 = 8×10-12 — (1)
Case II
When the two identical spheres are joined with a wire then the charges will be distributed equally on both of them.
New Charges will be
q1'=q2'=(q1+q2)/2
F= kq1q2/r²
Fr²/k = q1q2
q1q2 = Fr²/k
q1q2 = [(8×10-3)×(3)²]/9×109
q1q2 = 8×10-12 — (1)
Case II
When the two identical spheres are joined with a wire then the charges will be distributed equally on both of them.
New Charges will be
q1'=q2'=(q1+q2)/2
F' = 9 mN = 9×10-3 N
Again Using Coulomb's Law,
F' = kq1'q2'/r²
F'r²/k = q1'q2'
q1'q2' = F'r²/k
Again Using Coulomb's Law,
F' = kq1'q2'/r²
F'r²/k = q1'q2'
q1'q2' = F'r²/k
q1'q2' = [9×10-3×(3)²]/(9×109)
[(q1+q2)/2][(q1+q2)/2] = 9×10-12
(q1+q2)²/4 = 9×10-12
(q1+q2)² = 36×10-12 — (2)
q1+q2 = 6×10-6 — (3)
Now, Using the relation
(q1-q2)² = (q1+q2)²-4q1q2
(q1-q2)² = 36×10-12-4[8×10-12]
(q1-q2)² = 36×10-12-32×10-12
(q1-q2)² = 4×10-12
q1-q2 = 2×10-6 — (4)
Adding equation (3) and (4)
q1+q2+q1-q2 = (6+2)×10-6
2q1 = 8×10-6
q1 = (8/2)×10-6
q1 = 4×10-6 C
Putting in eq (3)
4×10-6 +q2 = 6×10-6
q2=(6-4)×10-6
q2=2×10-6 C
Hence, q1=4µC and q2=2µC
[(q1+q2)/2][(q1+q2)/2] = 9×10-12
(q1+q2)²/4 = 9×10-12
(q1+q2)² = 36×10-12 — (2)
q1+q2 = 6×10-6 — (3)
Now, Using the relation
(q1-q2)² = (q1+q2)²-4q1q2
(q1-q2)² = 36×10-12-4[8×10-12]
(q1-q2)² = 36×10-12-32×10-12
(q1-q2)² = 4×10-12
q1-q2 = 2×10-6 — (4)
Adding equation (3) and (4)
q1+q2+q1-q2 = (6+2)×10-6
2q1 = 8×10-6
q1 = (8/2)×10-6
q1 = 4×10-6 C
Putting in eq (3)
4×10-6 +q2 = 6×10-6
q2=(6-4)×10-6
q2=2×10-6 C
Hence, q1=4µC and q2=2µC
