A positive charge +q1 is located to the left of a negative charge −q2 On a line passing through the two charges, there are two places where the total potential is zero. The reference is assumed to be at infinity The first place is between the charges and is 4.80 cm to the left of the negative charge. The second place is 7.35 cm to the right of the negative charge if q2=−12μC , what is the value of charge q1∈μC?

Explanation:

At the point where net potential is zero , magnitude of potential will be equal .

At point between the given charges potential can be expressed as follows

k q₁ / d x 10⁻² = k 12 x 10⁻⁶ / 4.8 x 10⁻² ( q₁  charge is at distance d from the point )

At other point  potential due to both the charges  can be expressed as follows

k q₁ / (d+12.15) x 10⁻²  = k 12 x 10⁻⁶ / 7.35 x 10⁻²

(d+12.15) / d = 7.35 / 4.8

4.8 d + 58.32 = 7.35 d

2.55 d = 58.32

d = 22.87

Putting this value in first equation

k q₁ / d x 10⁻² = k 12 x 10⁻⁶ / 4.8 x 10⁻²

q₁ / 22.87 =  12 x 10⁻⁶ / 4.8

q₁ = 57.17 x 10⁻⁶

=  57.17 μC

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft