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

From the principle of conservation of linear momentum
   Force x distance1 = opposite force x distance 2
 165 x 0.4 = 0.45 x F2
 F2= 165x0.4/0.45
 F2= 147N
 The second child must exert a force of 147N to keep the door from moving.