Question

Question 2 6.25 pts For distances z which are large compared to the charge separation d, the electric field of an electric dipole is given by Ē 2kqd ŷ. If the charge is a 23 = 4.77 mC, and the dipole separation is d = 1.33 um, what is the magnitude of the electric potential of the dipole at z = 28.4 cm, assuming the potential is zero when one is infinitely far away from the dipole. Z -9 d +9

Answer 1

2) Given:

q = 4.77 mC
d = 1.33 micro-meter

z = 28.4 cm.

The potential due to a dipole at an axis point is $V = \frac{kqd}{r^{2}}$ [where r = distance from the center of the dipole and is much greater than d]

Now, in our problem, instead of 'r', z is used.

therefore, the potential at the point of distance z = 28.4 cm is $V = \frac{kqd}{z^{2}} = \frac{9*10^{9}*4.77*10^{-3}*1.33*10^{-6}}{(28.4*10^{-2})^{2}}=707.91 V$ [answer]

9) The relation between uniform electric field and an electric potential is $\Delta V = -Er$ [where r is the separation between two equipotential surfaces in the direction of the electric field]

For conducting plates, the electric field is $E=\frac{\sigma}{\epsilon_{o}}=\frac{6.66*10^{-9}}{8.85*10^{-12}}N/C$

Therefore, the potential difference between the two points of separation r (= 4.44 - 1.11 = 3.33 cm) is

$\Delta V = -E*r=-\frac{6.66*10^{-9}}{8.85*10^{-12}}*3.33*10^{-2}=-25.06V$

Therefore, the magnitude of potential difference is at the separation 3.33 cm is 25.06 V. [answer]