QR long rod of length L and charge QR and a point charge Opc are fixed in ce near point P. Point P is a distance d from both the point charge d from the center of the rod. The electric field at point P is at an angle = 45°, as shown in the figure. ) How do the signs and magnitudes of QR and Opc compare? Explain your reasoning qualitatively (without math). L d P Determine an explicit, quantitative (with math) expression for the charge on the rod QR in terms of the charge of the point charge Opc. Is this expression consistent with your answer above? How do you know? Show your work and explain your reasoning.
QR long rod of length L and charge QR and a point charge Opc are fixed in ce near point P. Point P is a distance d from both the point charge d from the center of the rod. The electric field at point P is at an angle = 45°, as shown in the figure. ) How do the signs and magnitudes of QR and Opc compare? Explain your reasoning qualitatively (without math). L d P Determine an explicit, quantitative (with math) expression for the charge on the rod QR in terms of the charge of the point charge Opc. Is this expression consistent with your answer above? How do you know? Show your work and explain your reasoning.
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Question
please answer all parts. First image is equation sheet
Transcribed Image Text:QR
A long rod of length L and charge QR and a point charge Opc are fixed in
place near point P. Point P is a distance d from both the point charge
and from the center of the rod. The electric field at point P is at an angle
0 = 45°, as shown in the figure.
a)
How do the signs and magnitudes of QR and Ope
compare? Explain your reasoning qualitatively (without math).
L
d
P
2pc
45°
b)
Determine an explicit, quantitative (with math) expression for the charge on the rod
QR in terms of the charge of the point charge Opc. Is this expression consistent with your
answer above? How do you know? Show your work and explain your reasoning.
Transcribed Image Text:|Epc|=
W = F - Ax = |F||Ax| cos
Mathematical Relations:
√2
sin (0)
kQ
p²
45°
1
45°
0
H
kQq
|²| = =
p2
1
| Ēroa| = 22
2kQ
Erod
rL
ΔΡΕ =
Vpc
=
cos(0)
=qAV
F = qË
EWext
kQ
A
H
H² = A² + 0²
30⁰
==▬▬▬▬
2kλ
√3/2
r
|Ē| :
= AE = AKE + APE + ··
tan (0)
=
=
Vsheet = Ex
60°
p = Qd
0
A
|Esheet | =
=
AV
Ax
Special Triangles:
Note that you can use these to determine the sine/cosine of the angles, e.g. cos(45) = 1/√2 = √2/2
1/2
2лkQ
A
= 2лko
0
5
37⁰
Wint = -APE
H
4
A
53°
3
Expert Solution
Step 1
For any resultant vector which is angle bisector then magnitude of both vector must be equal
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