Physics For Scientists And Engineers: A Strategic Approach, 4th Edition
Authors: Randall D. Knight
ISBN-13: 978-0133942651
See our solution for Question 67P from Chapter 22 from Randall Knight's Physics for Scientists and Engineers.
Problem 67P
Chapter:
Problem:
...causes the 5.0 g ball in Figure P20.65 to hang at a 20° angle. What is the charge on the ball?
Step-by-Step Solution
Step 1
We are given that a charged (q) ball hangs with a string in an electric field (E) of 100000 N/C in x direction. The weight of the ball (m) is 50 g. There are two forces acting on the ball, the electric field causes movement in x direction whereas the horizontal component of the gravitational force counter-balances the electric field.
Step 2: The Free body diagram of the ball
Step 3: Equilibrium of forces
Electric Force: ${F_e} = qE$
Gravitational Force: ${F_g} = mg$
Equilibrium of forces in vertical direction:\[\begin{array}{l}T\cos \theta = {F_g}\\T\cos \theta = mg\,\,\,\,....\left( 1 \right)\end{array}\]Equilibrium of forces in horizontal direction:\[\begin{array}{l}T\sin \theta = {F_e}\\T\sin \theta = qE\,\,\,\,....\left( 2 \right)\end{array}\]
Step 4: The charge on the ball
Divide equation (1) and (2)\[\begin{array}{l}\dfrac{{T\sin \theta }}{{T\cos \theta }} = \dfrac{{qE}}{{mg}}\\\tan \theta = \dfrac{{qE}}{{mg}}\\q = \dfrac{{mg\tan \theta }}{E}\end{array}\]Substitute the data: \[\begin{array}{l}q = \dfrac{{mg\tan \theta }}{E}\\\\ = \dfrac{{\left( {\dfrac{5}{{1000}}\,\,{\rm{kg}}} \right)\left( {9.8\,\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\tan \left( {20^\circ } \right)}}{{100,000\,\,{\rm{N/C}}}}\\ = 178.345414 \times {10^{ - 9}}\,\,C\\ = 178.345414\,\,{\rm{n}}C\\ \approx 178\,\,{\rm{n}}C\end{array}\]
ANSWERS
The charge on the ball is: \[q \approx 178\,\,{\rm{n}}C\]
We are given that a charged (q) ball hangs with a string in an electric field (E) of 100000 N/C in x direction. The weight of the ball (m) is 50 g. There are two forces acting on the ball, the electric field causes movement in x direction whereas the horizontal component of the gravitational force counter-balances the electric field.
Step 2: The Free body diagram of the ball
Step 3: Equilibrium of forces
Electric Force: ${F_e} = qE$
Gravitational Force: ${F_g} = mg$
Equilibrium of forces in vertical direction:\[\begin{array}{l}T\cos \theta = {F_g}\\T\cos \theta = mg\,\,\,\,....\left( 1 \right)\end{array}\]Equilibrium of forces in horizontal direction:\[\begin{array}{l}T\sin \theta = {F_e}\\T\sin \theta = qE\,\,\,\,....\left( 2 \right)\end{array}\]
Step 4: The charge on the ball
Divide equation (1) and (2)\[\begin{array}{l}\dfrac{{T\sin \theta }}{{T\cos \theta }} = \dfrac{{qE}}{{mg}}\\\tan \theta = \dfrac{{qE}}{{mg}}\\q = \dfrac{{mg\tan \theta }}{E}\end{array}\]Substitute the data: \[\begin{array}{l}q = \dfrac{{mg\tan \theta }}{E}\\\\ = \dfrac{{\left( {\dfrac{5}{{1000}}\,\,{\rm{kg}}} \right)\left( {9.8\,\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\tan \left( {20^\circ } \right)}}{{100,000\,\,{\rm{N/C}}}}\\ = 178.345414 \times {10^{ - 9}}\,\,C\\ = 178.345414\,\,{\rm{n}}C\\ \approx 178\,\,{\rm{n}}C\end{array}\]
ANSWERS
The charge on the ball is: \[q \approx 178\,\,{\rm{n}}C\]