University
Physics 2
Physics 2
Spring 2025
Three conducting spheres, A, B and C, have charges \(Q_A = +16\) μC, \(Q_B = -4\) μC, and \(Q_C = 0\) μC, respectively.
Determine \(Q_A\), \(Q_B\) and \(Q_C\) after the following maneuvers. Each maneuver begins with the initial charges given above.
The electroscope in figure a is uncharged.
In figure b, a positively charged rod approaches but does not touch, and the electroscope leaves separate.
In figure c, the electroscope is grounded while the rod remains in place (you will determine the position of its leaves).
In figure d, the ground is removed, then the rod is removed.
In each case, sketch the electroscope and the charge distribution on it. For c and d, also sketch the position of the leaves.
a.
b.
c.
d.
a.
b.
c.
d.
The two charges shown have charge \(+1e\) and \(-4e\).
Sketch the direction of the electric field at the labeled points, or state \(0\) for zero field. (You may ignore the magnitude of the vectors.)
a.
b.
c.
d.
e.
f.
Point charges \(Q_1=2.0\) μC and \(Q_2=4.0\) μC are located at \(\vec{r}_1 = (4.0 \hat{\imath} - 2.0 \hat{\jmath} + 5.0 \hat{k})\) m and \(\vec{r}_2 = (8.0 \hat{\imath} + 5.0 \hat{\jmath} - 9.0 \hat{k})\) m.
What is the force by \(Q_2\) on \(Q_1\)?
Express your answer as a vector in \(\hat{\imath} \; \hat{\jmath} \; \hat{k}\) notation.
We first need to find the vector that points from \(Q_2\) to \(Q_1\) (call it \(\vec{r}_{21}\)). To get this, imagine the vector sum going from the origin to \(Q_2\) then from \(Q_2\) to \(Q_1\), this would end up at \(Q_1\). So:
\(\vec{r}_{2} + \vec{r}_{21} = \vec{r}_{1}\)
This means that
\(\vec{r}_{21} = \vec{r}_{1} - \vec{r}_{2} = (-4.0 \hat{\imath} - 7.0 \hat{\jmath} + 14.0 \hat{k})\) m.
The magnitude of this vector is \(r_{21} = 16.16\) m. So the unit vector pointing from \(Q_2\) to \(Q_1\) is
\(\hat{r}_{21} = (-0.248 \hat{\imath} - 0.433 \hat{\jmath} + 0.867 \hat{k})\).
The magnitude of the force is
\(F = 8.99 \times 10^9 \cdot \frac{8.0 \times 10^{-12}}{16.16^2} = 2.75 \times 10^{-4}\) N.
So the force vector is
\(\vec{F} = 2.75 \times 10^{-4} (-0.248 \hat{\imath} - 0.433 \hat{\jmath} + 0.867 \hat{k})\) N.
\(\vec{F} = (-6.82 \times 10^{-5}) \hat{\imath} - (1.19 \times 10^{-4}) \hat{\jmath} + (2.39 \times 10^{-4}) \hat{k}\) N.
A small sphere of mass \(0.25\) g that carries a charge of \(9.0 \times 10^{−10}\) C. The sphere is attached to one end of a very thin silk string 5.0 cm long. The other end of the string is attached to a large vertical conducting plate that has a charge density of \(30 \times 10^{−6}\) C/m². What is the angle that the string makes with the vertical?
The magnitude of the electric field due to the wall’s surface charge is
\(E = \frac{\sigma}{2 \epsilon_0} = 1.70 \times 10^6\) N/C
with direction to the right.
The force on the sphere is to the right with magnitude
\(F = qE = 1.53 \times 10^{-3}\) N.
A free-body diagram of the sphere shows the three forces (gravity \(mg\), tension \(T\) and electric repulsion \(qE\)) must balance, with force sums as follows:
\(\sum F_x = 0 = qE - T \sin \theta\)
\(\sum F_y = 0 = T \cos \theta - mg\)
Eliminate \(T\) from these equations to obtain
\(\tan \theta = \frac{qE}{mg} = 0.62\)
So the angle is
\(\theta = 32^{\circ}\)
Three charges are positioned at the corners of a parallelogram as shown below.
Express your answers in \(\hat{\imath} \; \hat{\jmath} \; \hat{k}\) notation.
Define \(\vec{r}_1\), \(\vec{r}_2\) and \(\vec{r}_3\) to point from the respective charge to the unoccupied corner. Then
\(\vec{r}_3 = (3.0 \hat{\imath})\) m,
\(\vec{r}_2 = ( \frac{1}{\tan 30^{\circ}} \hat{\imath} + 1.0 \hat{\jmath})\) m
\(\vec{r}_1 = \vec{r}_2 + \vec{r}_3\)
Add the electric field of each.
\(\vec{E} = (10.2 \hat{\imath} + 18.6 \hat{\jmath})\) kN/C
\(\vec{F} = (50.9 \hat{\imath} + 93.1 \hat{\jmath})\) mN
The charge per unit length on the thin semicircular wire shown below is \(\lambda\). What is the electric force on the point charge \(q\)?
Define the \(xy\) axes. It is convenient if the charge \(q\) is at the origin, with \(+x\) to the right. and \(+y\) upward. The electric field at the origin due to the wire is found (below) to be
\(\displaystyle \vec{E} = - \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{R} \hat{\jmath}\)
With this field the solution is
\(\displaystyle \vec{F} = q \vec{E} = - \frac{q}{2 \pi \epsilon_0} \frac{\lambda}{R} \hat{\jmath}\)
To find the electric field due to the wire:
\(\displaystyle \vec{E} = \int d\vec{E} = \frac{1}{4 \pi \epsilon_0} \int \frac{dq}{r^2} \hat{r}\)
The \(r^2\) denominator is the constant \(R^2\).
The unit vector changes over the integration. To get an expression for \(\hat{r}\), define \(\theta\) as the angle off vertical, so the integral will have limits from \(\theta = -\frac{\pi}{2}\) to \(\theta = \frac{\pi}{2}\). Then \(\hat{r}\) (the unit vector pointing from \(dq\) toward the origin) can be written
\(\hat{r} = -\hat{\imath} \sin \theta - \hat{\jmath} \cos \theta\)
(note the negative \(\hat{\imath}\) component is due to the fact that negative angles require a positive \(x\) component).
Now the integral is
\[ \vec{E} = - \frac{1}{4 \pi \epsilon_0} \frac{1}{R^2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\hat{\imath} \sin \theta + \hat{\jmath} \cos \theta ) dq\]
Before we integrate we require \(dq\) to be written in terms of the variable \(\theta\): \(dq = \lambda R d\theta\).
\[ \vec{E} = - \frac{1}{4 \pi \epsilon_0} \frac{\lambda}{R} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\hat{\imath} \sin \theta + \hat{\jmath} \cos \theta ) \; d\theta\]
Each term can be integrated separately.
\[\vec{E} = - \frac{1}{4 \pi \epsilon_0} \frac{\lambda}{R} \left ( \hat{\imath} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin \theta \; d\theta + \hat{\jmath} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos \theta \; d\theta \right ) \]
Evaluating the integral (the \(\hat{\imath}\) term will integrate to zero):
\[\vec{E} = - \frac{1}{4 \pi \epsilon_0} \frac{\lambda}{R} \left ( 0 + \hat{\jmath} \left (\sin \theta) \right |_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \right ) = - \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{R} \hat{\jmath} \]
Last modified: Mon February 10 2025, 05:01 PM.