A 7.1 nC point charge and a - 2.0 nC point charge are 3.7 cm apart. What is the electric field strength at the midpoint between the two charges

2.39 * 10^5 N/C

Explanation:

Given that:

q1 = 7.1 nC = 7.1 * 10^-9 C

q2 = - 2.0 nC = 2 * 10^-9 C

Distance (d) = 3.7 cm apart

Hence, midpoint (r) = 3.7/2 = 1.85 cm = 1.85/100 =0.0185 m

The electric field strength at midpoint is given by:

E = k/r² ( q1 + q2)

K = 9.0 * 10^9

E = ((9 * 10^9) / 0.0185^2) * (7.1 * 10^-9 + 2 * 10^-9)

E = (2.62965 * 10^13) * (9.1 * 10^-9)

E = 23.929815 * 10^13 - 9

E = 23.929815 * 10^4

E = 2.39 * 10^5 N/C

i think the answer would be reserve.

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