Question

Each of two small spheres is charged positively, the combined charge being 7.40E-05 C. If each sphere is repelled from the other by a force of 3.50 N. when the spheres are 0.30 m. apart, what is the smaller charge (in coulombs)?

Answer 1

ANSWER :

Given data : Combined charge Q= (Q1+Q2) = 7.40 E^-05 C   -------------------------------- (1)
                   REPELLING FORCE (F) = 3.50 N
                   Sphere separation (r) = 0.30 m

we know from Coulomb's Law: F = (k Q ₁ x Q ₂₎ / r ² , where k is a constant
                                   k = = 9.0 x 10 ⁹ N-m²⁄C²

Rearranging the equation for Coulomb's Law , we have:

Q ₁ x Q ₂ = (F x r²)/k = ( 3.50 x 0.30 x 0.30 ) / 9.0 x 10 ⁹ = 3.5 x 10 ^-11 C²   ----------------------- (2)

From the Algebra equation (a-b)2 = (a+b)2 - 4ab   using it for the above case, we can re write it as

(Q1-Q2)2 = (Q1+Q2)2 - 4Q1 Q2

(Q1-Q2)2 = (7.40 E^-05)2 - 4(3.5 x 10 ^-11 )

(Q1-Q2) = 7.3048 E^-05 C    ------------------------------------ (3)

SO, solving (1) & (3) , we have Q1 = ~ 7.35 E^-05     & Q2 = ~ 0.05 E^-05

So, Final ANSWER FOR THE QUESTION IS : SMALLER CHARGE IS approximately 0.05 E^-05 C