Question

Three charges A, B and C are placed in a straight line and in this order. The charge of A is twice that of C and has a value of 40.0nC. B carries 6.25 x 10^11 electrons. A is 5cm from B and 8.5cm from C.
(a) Show that the charge of B is -100nC.
(b) Calculate the resultant force exerted on C due to A and B.
(c) Show without any calculation, that the resultant force on C cannot be zero if it is now placed between A and B.

Answer 1

1. To calculate the charge of B we just need to multiply an electron charge by the quantity of electrons : $q_B = N \cdot e = 6.25\cdot10^{11}\cdot(-1.6)\cdot10^{-19} = -10^{-7}C = -100nC$
2. To calculate this force we just apply directly the Coulomb's law to couples of charges A,C and B,C and we add up the results : $F= \frac{1}{4\pi \varepsilon_0} (\frac{q_A q_c}{|AC|^2} + \frac{q_Bq_C}{|BC|^2})$ and by using the numerical values we have $F = \frac{1}{4\cdot\pi\cdot 8.85\cdot10^{-11}}\cdot(\frac{40\cdot 20\cdot10^{-18}}{8.5^2\cdot10^{-4}}-\frac{20\cdot100\cdot10^{-18}}{3.5^2\cdot10^{-4}})\approx -13.7 \cdot10^{-3} N = -13.7N$ and the sign - means that the charged C is pulled towards the direction of charges B and A.
3. If we place the charge C between A and B, it will be repulsed by a non-zero force from the direction of the charge A (as they have the charges of the same sign) and it will be pulled by a non-zero force towards the charge B (as their charge signs are opposite). But as these directions coincide in this configuration, the resultant force (directed towards B) is the sum of two positive quantities and thus non-zero.