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Electrostatics &
Electric Potential
5 Sample Problems
1. Point Chargesβ’ Find the Electric Field @ point P
caused by a bunch of individual charges.
E is a vector, so direction is important! Break the Eβs into components if necessary.
+
+-
P
β’ Find the Electric Potential @ point P caused by a bunch of individual charges.
VV is a scalar, so direction isnβt important; however, it is possible to have a negative value, so donβt ignore the sign.
2. Rod (E)β’ Find the Electric Field
@ point P.
1. 2. ππ=πππ₯
3.πΈ=β« ππππ₯π₯2 =π πβ«
π
π+πΏ
π₯β2ππ₯
P a
L
1. Equation
2. Since r and q change together, we need an equation that relates the two.Charge Density. The density stays constant whether over the total charge or parts of the rods.
3. Solve for dq and substitute it in. Pull your constants out of the integral.Determine the range and place it on the integral.
2. Rod (E)β’ Find the Electric Field
@ point P.
P a
L
4 .πΈ=π π (β1 )π₯β1
a
a+L
6.
4. Integrate.
6. Plug in the ranges and simplify.
5 .πΈ=πππΏ
(β1 )π₯β1
a
a+L 5. Substitute in the total charge density.
2. Rod (V)β’ Find the Electric
Potential @ point P.
1. 2. ππ=πππ₯
3.π=β« π πππ₯π₯
=π π β«π
π+πΏ
π₯β1ππ₯
P a
L
1. Equation
2. Since r and q change together, we need an equation that relates the two.
Charge Density. The density stays constant whether over the total charge or parts of the rods.
3. Solve for dq and substitute it in. Pull your constants out of the integral.
Determine the range and place it on the integral.
2. Rod (V)β’ Find the Electric Field
@ point P.
P a
L
4 .π=π ππππ₯a
a+L
6.
4. Integrate.
6. Plug in the ranges and simplify.
5 .π=πππΏπππ₯
a
a+L 5. Substitute in the total charge density.
Notice that the steps for solving were the same in both cases.
In all the different problems, those steps stay the same. The only thing that changes is how you do the step. For instance, how you integrate, or whether you use linear charge density, area charge density, or volume charge density, etc.
3. Ring (E)β’ Find the Electric Field
@ point P.
1. 2. 3.πΈ=β« πππ
(πΒΏΒΏ2+π₯2)π₯
β(π2+π₯2)
ΒΏ
1. Equation
2. Donβt need charge density because all the charges are equidistant adding the same values to the field.or a & x are constant.
3. Replace dq. Pull out constants. Set Range.
πΈ=ππ₯
(πΒΏΒΏ 2+π₯2)β 3
2β«ππ ΒΏ
x
a P
3. Ring (E)β’ Find the Electric Field
@ point P.
4 .πΈ=ππ₯
(πΒΏΒΏ2+π₯2)β 3
2π ΒΏ4. Integrate.
6. No range since you will probably know the total charge.
5. No charge density to substitute!
x
a P
3. Ring (V)β’ Find the Electric
Potential @ point P.
1. 1. Equation2. a & x are constants!
3.π=π
(πΒΏΒΏ2+π₯2)β«ππ ΒΏ3. Replace dq. Pull out constants. Set Range.
4.πΈ=π
(πΒΏΒΏ2+π₯2)πΒΏ
4. Integrate.
6. No range since you will probably know the total charge.
5. No charge density to substitute!
x
a P
4. Disk (E)β’ Find the Electric Field
@ point P.
1.
2. ππ΄=2ππππ
3.πΈ=β« ππ₯π ππ΄
(π ΒΏΒΏ2+π₯2)32
=β« ππ π₯2π πππ
(π ΒΏΒΏ2+π₯2)32=ππ π π₯β«
0
π 2πππ
(π ΒΏΒΏ2+π₯2)32
ΒΏΒΏ ΒΏ
1. Equation
2. Charge Density. Area because weβre moving out concentric circles.
3. Replace dq. Pull out constants. Set Range.
x
r P
4. Disk (E)β’ Find the Electric Field
@ point P.
4. Integrate.
6. Plug in the ranges and simplify.
5 .πΈ=β2πππ π₯1
(π 2+π₯2)12 0
R5. Substitute in the
total charge density.
4.πΈ=πππ π₯β«0
π ππ
π32
=πππ π₯π
β 12
β12
=πππ π₯π
β 12
β12
=β2πππ π₯1
π12
0
R
6 .πΈ=2πππ [1β π₯
(π2+π₯2)12 ]
x
r P
4. Disk (V)β’ Find the Electric
Potential @ point P.
3.π=β« ππ ππ΄
βπ2+π₯2=β« ππ 2ππππ
βπ 2+π₯2=ππ2πβ«
0
π πππ
βπ2+π₯2
1. Equation
2. Charge Density: Area Again
1. 2. ππ΄=2ππππ
3. Replace dq. Pull out constants. Set Range.
x
r P
4. Disk(V)β’ Find the Electric Field
@ point P.
4. Integrate.
6. Plug in the ranges and simplify.6 .π=2πππ (β(π 2+π₯2) β x )
4 .π=ππ 2πβ«0
π πππ
βπ 2+π₯2=ππ 2πβ«
0
π πππ2π
π12
=πππβ«0
π ππ
π12
=ππππ
12
12
0
RΒΏ2πππ β(π2+π₯2)
x
r P
5. Arc Length (E)β’ Find the Electric Field
@ point P.
1.
2. ππ=πππ
3.πΈ=β« πππππ2 πππ π=
π ππ2 β«
π1
π2
πππ π ππ
1. Equation (cos because all the y values cancel out.)
2. Charge Density.
πΈ=π ππ2 β«
π1
π2
πππ πππ π=π ππ β«
π 1
π 2
πππ πππ
ΞΈ changes with respect to the arc length!
3. Replace dq. Pull out constants (r is constant!). Set Range.
x
rPπ
5. Arc Length (E)β’ Find the Electric Field
@ point P.
6.
4. Integrate.
6. Plug in the ranges and simplify.
5 .πΈ=πππΏπ πππ
π1
π2 5. Substitute in the total charge density.
π1
π24.πΈ=
π ππ
π πππ
7. 7. The length is the arc length.
x
rPπ
5. Arc Length (V)β’ Find the Electric Field
@ point P.
1. 2. ππ=πππ
3.π=β« π ππππ
=π ππ β«
π1
π2
πππ=π πβ«π1
π2
ππ
1. Equation
2. Charge Density.
3. Replace dq. Pull out constants (r is constant!). Set Range.ΞΈ changes with respect to the arc length! Must be in Radians though (since ΞΈ is not in a function.)
x
rPπ
5. Arc Length (V)β’ Find the Electric Field
@ point P.
6.
4. Integrate.
6. Plug in the ranges and simplify.
5 .πΈ=πππΏππ1
π2 5. Substitute in the total charge density.
π1
π24.π=π ππ
7. 7. The length is the arc length.
x
rPπ