- Download PDF
Question on point charge particles in a square
Two pos. and two neg. point particles, all with charge of magnitude q = 5nC, are placed at the 4 corners of a 10 cm-side square as shown. (http://tinypic.com/r/2zrgnwi/6)
(a) Determine the electric ﬁeld, magnitude and direction, at the center of the square.
(b) Determine the energy required to assemble these four point charges from far apart to the present conﬁguration.
(c) An electron (me = 9.11 x 10^-31 kg) is placed at the center of the square with zero initial velocity. Determine its instantaneous acceleration, both direction and magnitude, due to the four point charges at the corners.
Any help would greatly be appreciated!
1 Answer | Add Yours
Since we are dealing with a test charge at the center of a square of side 0.1 m, the field from each corner will be directed along a diagonal of the square, toward the negative charges and away from the positive charges by convention.
If E is the magnitude of the field from one corner, the resultant charge will be `E_\text(net)=2sqrt(2)*E` directed up the positive y-axis. The E from the opposite corners will add together, which will result in two 2E vectors at right angles to each other. Pythagorean theorem does the rest. `r = 5*sqrt(2) \text(cm)` .
Since `E= (kq)/r^2` where k = 9\times10^9 \text(Nm^2)/\text(C^2)
gives us `E = 9000 \text(N)/\text(C)` and `E_\text(net)=25500 \text(N)/\text(C)`
Part c asks us to find the acceelration of an electron placed at the origin. Since `q_e = 1.602\times10^-19 \text(C) and m=9.11\times10^-31 \text(kg)`
then `a = (qE)/m`
Working through the math, we get `a = 4.48\times10^15 \text(m)/\text(s^2)`
The work done to get each charge together is derived from the equation for potential energy. We build one corner at a time. Going counter-clockwise from the top left corner. But first, we will determine the magnitudes of the possible potential energies associated with a side and a diagonal. Then determine whether it is positive or negative. Then add up all of the energies for the net result.
It takes 0 Work to place a charge when there is no field.
Placing a charge along one of the sides requires a change in potential energy of:
`U=(kQq)/r` where `r = 0.10 \text(m)`
Which comes to:
`U = 2.25\times10^-6 \text(J)`
Placing a charge diagonally across from one other charge in this case requires a change in potential energy of
`U=(kQq)/r` where `r = 0.10sqrt(2) \text(m)=0.141\text(m)`
Which comes to
`U = 1.59\times10^-6 \text(J)`
If we place the first charge -q, total energy is 0.
Then we place the next charge, -q the work required is `+2.25\muJ`
Now we place the third charge, +q, the work required is `-2.25\muJ` from the side charge above it, and `-1.59\muJ` from the diagonal charge. Making a total energy so far of `-1.59\muJ`
Now place the fourth charge, +q, the energy due to the other positive charge is `+2.25\muJ` , the -q directly above is `-2.25\muJ` , and due to the -q across is `-1.59\muJ` .
Thus the total work required is `-3.18\muJ`
We’ve answered 319,471 questions. We can answer yours, too.Ask a question