Question

Course Contents Prol Print emas 1 » 18.26 Louiomb U Notes Evaluate Feedback Two point charges Q1 and Q2 are 1.90 m apart, and their total charge is 19.2 PC. If the force of repulsion between them is 0.214 N, what are magnitudes of the two charges? Enter the smaller charge in the first box. Q1 = 12.10uc Q2 = 7.100C Submit Answer Incorrect. Tries 3/5 Previous Tries If one charge attracts the other with a force of 0.163N, what are the magnitudes of the two charges if their total charge is also 19.2 PC? The charges are at a distance of 1.90 m apart. Note that you may need to solve a quadratic equation to reach your answer. Enter the charge with a smaller magnitude in the first box. Q1 = -3.00uC Q2 = 22.15uc Submit Answer Incorrect. Tries 3/5 Previous Tries This discussion is closed. Send Feedback

Answer 1

1)

Given:

Distance between the charges,

T= 1.90m

Total charge,

Οι + O2 19.2μα

${\color{Red} Q_{2}=19.2-Q_{1}}$-------------(1)

Repulsive force between the charges,

F = 0.214N

Electric force, $F=\frac{kQ_{1}Q_{2}}{r^{2}}$

$0.214N=\frac{9*10^{9}*Q_{1}Q_{2}}{(1.90m)^{2}}$

9 * 10°* Q1Q2 0.214 1.92

$\frac{0.214*1.9^{2}}{9*10^{9}}=Q_{1}Q_{2}$

8.5838 + 10" = 2182

$85.838*10^{-12}C^{2}=Q_{1}Q_{2}$

$85.838\mu C^{2}=Q_{1}Q_{2}$

Put (1) in the above equation

85.838 C2 = Q1(19.2°C - Q1)

$85.838=Q_{1}(19.2-Q_{1})$

$85.838=19.2Q_{1}-Q_{1}^{2}$

$Q_{1}^{2}-19.2Q_{1}+85.838=0$

Solve the quadratic equation using a calculator.

ANSWER: ${\color{Red} Q_{1}=7.086\mu C}$

Put the value of Q₁ in (1)

$Q_{2}=19.2\mu C-7.086\mu C$

ANSWER: ${\color{Red} Q_{2}=12.114\mu C}$

Note: It is specified in the question to put the smaller charge in the first box

==========================================================

2)

Given:

Distance between the charges,

T= 1.90m

Total charge,

Οι + O2 19.2μα

${\color{Red} Q_{2}=19.2-Q_{1}}$-------------(2)

Attractive force between the charges, $F=-0.163N$

Note: In the 1st case we took the repulsive force as positive, so take the attractive force as negative.

Electric force, $F=\frac{kQ_{1}Q_{2}}{r^{2}}$

$-0.163N=\frac{9*10^{9}*Q_{1}Q_{2}}{(1.90m)^{2}}$

$-0.163=\frac{9*10^{9}*Q_{1}Q_{2}}{1.9^{2}}$

$-\frac{0.163*1.9^{2}}{9*10^{9}}=Q_{1}Q_{2}$

$-6.5381*10^{-11}=Q_{1}Q_{2}$

$-65.381*10^{-12}C^{2}=Q_{1}Q_{2}$

$-65.381\mu C^{2}=Q_{1}Q_{2}$

Put (2) in the above equation

$-65.381\mu C^{2}=Q_{1}(19.2\mu C-Q_{1})$

$-65.381=Q_{1}(19.2-Q_{1})$

$-65.381=19.2Q_{1}-Q_{1}^{2}$

$Q_{1}^{2}-19.2Q_{1}-65.381=0$

Solve the quadratic equation using a calculator.

ANSWER: ${\color{Red} Q_{1}=-2.952\mu C}$

Put the value of Q₁ in (2)

$Q_{2}=19.2-Q_{1}$

$Q_{2}=19.2\mu C--2.952\mu C$

$Q_{2}=19.2\mu C+2.952\mu C$

ANSWER: ${\color{Red} Q_{2}=22.152\mu C}$

Note: It is specified in the question to put the smaller charge in the first box

==========================================================