Question

cswebdav/pd-203 1099-dt-content-rid-127647332/courses/201 920-PHYS 1402010/Homework%281962 PHYS 1402 (General Physics II) Homework 2 1. When we defined the electric potential, we said that we take a point in- finitely far away to have a potential of zero when dealing with a finite distribution of charges. Why might this be a bad approximation for a non-finite distribution of charges? 2. Consider a dipole centered at the origin of a cartesian coordinate system. The dipole is aligned with the x-axis with the positive charge on the pos- itive axis. The magnitude of each charge is (q) and there is a distance of (d) between the dipole charges. Find the torque on the dipole if an electric field, with magnitude (E), permeates the region of the dipole if the electric field a) points along the x-axis b) points along the y-axis, c) points at a 45 degree angle between the x and y axes. LG

Answer 1

2.

a)

E = magnitude of electric field along positive x-direction

$\underset{E}{\rightarrow}$= E i

p = magnitude of dipole moment = q d

$\underset{P}{\rightarrow}$= p i = qd i

Torque is given as

$\tau$ = $\underset{P}{\rightarrow}$ x $\underset{E}{\rightarrow}$

$\tau$ = (qd) i x (E) i

$\tau$ = (qdE) (i x i )

$\tau$ = 0 Nm

b)

E = magnitude of electric field along positive j -direction

$\underset{E}{\rightarrow}$= E j

p = magnitude of dipole moment = q d

$\underset{P}{\rightarrow}$= p i = qd i

Torque is given as

$\tau$ = $\underset{P}{\rightarrow}$ x $\underset{E}{\rightarrow}$

$\tau$ = (qd) i x (E) j

$\tau$ = (qdE) (i x j )

$\tau$ = (qdE) k

c)

E = magnitude of electric field

$\underset{E}{\rightarrow}$= (ECos45) i + (E Sin45) j

p = magnitude of dipole moment = q d

$\underset{P}{\rightarrow}$= p i = qd i

Torque is given as

$\tau$ = $\underset{P}{\rightarrow}$ x $\underset{E}{\rightarrow}$

$\tau$ = (qd) i x ((ECos45) i + (E Sin45) j)

$\tau$ = (qdE Sin45) (i x j )

$\tau$ = (qdE Sin45) k