Two charged spheres placed 43 cm apart exert a force of 1.40 × 10-14 N
on each other. If one of the spheres has a charge of 1.68 × 10-17 C,
what is the charge of the other sphere?
Answers
Answer:
The force between two point charges obeys the Coulomb's law:
F = k\dfrac{|q_1||q_2|}{r^2}F=k
r
2
∣q
1
∣∣q
2
∣
where q_1 = 1.68\times10^{-17}C,\space q_2q
1
=1.68×10
−17
C, q
2
are the charges of the first and second spheres respectively, r = 43cm = 0.43mr=43cm=0.43m is the distance between the spheres, and k = 9\times 10^9 N\cdot m^2/C^2k=9×10
9
N⋅m
2
/C
2
is the Coulomb's constant. Expressing the second charge and substituting F = 1.4\times 10^{-14}NF=1.4×10
−14
N , obtain:
q_2 = \dfrac{Fr^2}{kq_1}\\ q_2 = \dfrac{1.4\times 10^{-14}\cdot 0.43^2}{9\times 10^9\cdot 1.68\times 10^{-17}} \approx 1.7\times 10^{-10}Cq
2
=
kq
1
Fr
2
q
2
=
9×10
9
⋅1.68×10
−17
1.4×10
−14
⋅0.43
2
≈1.7×10
−10
C
Answer. 1.7\times 10^{-10}C1.7×10
−10
C.
Given:
Charge on one sphere is × C
The Force between the spheres ×
Distance between the spheres
To find:
The charge present on the other sphere.
Solution:
The force of attraction between two charged bodies as defined by Columb's Law is given by:
where is the charge present one first body and is the charge present on another body and is the distance of separation between the bodies
Substitute these values in the formula of force and calculated the value of
C
Thus, the charge on another sphere is C