Question

d

Point charges Q_1 = + 3 10^6 C and Q_2 = -9 10^-6 C are fixed on the x axis at x = -11 m and x = +11 m, respectively. Sketch the electric potential (voltage) on the x axis for this system. (NOT FOR CREDIT -- but essential for the next parts.) Your sketch should show that there is a point on the x axis between the two charges where V=0. Is this point closer to Q_1 or Q_2? Looking at your graph, is there an electric field at this point? Find the x coordinate of the point referred to in part b. m A +2 millicoulomb charge placed at rest at this point would have: a potential energy of: 3 (Anything significant about this value?) a kinetic energy of: 3 and it will not move When this charge passes through the origin, it will have: a potential energy of: a kinetic energy of:

Answer 1

a) b) The point is closer to Q1 as you can see in thr graph.

16 x=-5.6 V = 0.0

The electric field is given by the rate of change of the electric potential respect to the position. The electric field is zero if and only if the electric potential is constant. In the point the electric potential is clearely not constant, therefore the electric field at that point is not zero

c) The total electric potential at any point in the x axis is

$V(x)=k_e\frac{q_1}{|x+11|}-k_e\frac{q_2}{|x-11|}$

for the potential zero

$0=k_e\frac{q_1}{|x+11|}-k_e\frac{q_2}{|x-11|}$

$k_e\frac{q_2}{|x-11|}=k_e\frac{q_1}{|x+11|}$

${q_1}{|x-11|}={q_2}{|x+11|}$

Her we have to solve two equations

$q_1(x-11)=q_2(x+11)$

$q_1x-11q_1=q_2x+11q_2$

$-11q_1-11q_2=q_2x-q_1x$

$-11(q_1+q_2)=(q_2-q_1)x$

$x=-11\frac{q_1+q_2}{(q_2-q_1)}$

$x=-11\frac{3+9}{9-3}$

$x=22m$

but that point is not between the charges

the second equation is

$q_1(11-x)=q_2(x+11)$

$11q_1-q_1x=q_2x+11q_2$

$11q_1-11q_2=q_2x+q_1x$

$11(q_1-q_2)=(q_2+q_1)x$

$x=11\frac{q_1-q_2}{(q_2+q_1)}$

$x=11\frac{3-9}{9+3}$

$x=-5.5m$

That is the point

d) The potential energy like is not defined. But in electrostatic is common to reffere to this potential energy at one point the difference of potential energy from the infinite to the point. That potential energy is given by

$U=qV$for point charges.

Since the elecetric potential is zero at that point

$U=q(0)$

$U=0J$

The kinetic energy is

$K=\frac{1}{2}mv^2$

$K=\frac{1}{2}m(0)^2$

$K=0J$

The charge is pushed by the positive charge in +x direction and pulled by the negative charge in +x direction. The charge will moves to +x direction. Note that althought the potentail energy is zero, the charge is not in an equillibrium point.

e) At the origin the electrip potential is

$V(x)=(8.99\times 10^{9}N\cdot m^2/C^2)\frac{(3\times 10^{-6}C)}{|(0)+11m|}-(8.99\times 10^{9}N\cdot m^2/C^2)\frac{(9\times 10^{-6}C)}{|(0)-11m|}$$V(x)=-4900V$

Then the change in the potential energy is

$\Delta U=(2\times 10^{-3}C)(-4900V-0V)$

$\Delta U=-9.8J$

And by the conservation of the energy

$\Delta K=-\Delta U$

$(K_2-K_1)=-(-9.8J)$

$(K_2-(0J))=-(-9.8J)$

$K_2=9.8J$