Electrostatics &

Electric Potential

5 Sample Problems

1. Point Charges• Find the Electric Field @ point P

caused by a bunch of individual charges.

E is a vector, so direction is important! Break the E’s into components if necessary.

+

+-

P

• Find the Electric Potential @ point P caused by a bunch of individual charges.

VV is a scalar, so direction isn’t important; however, it is possible to have a negative value, so don’t ignore the sign.

2. Rod (E)• Find the Electric Field

@ point P.

1. 2. 𝑑𝑞=𝜆𝑑𝑥

3.𝐸=∫ 𝑘𝜆𝑑𝑥𝑥2 =𝑘 𝜆∫

𝑎

𝑎+𝐿

𝑥−2𝑑𝑥

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)• Find the Electric Field

@ point P.

P a

L

4 .𝐸=𝑘 𝜆 (−1 )𝑥−1

a

a+L

6.

4. Integrate.

6. Plug in the ranges and simplify.

5 .𝐸=𝑘𝑄𝐿

(−1 )𝑥−1

a

a+L 5. Substitute in the total charge density.

2. Rod (V)• Find the Electric

Potential @ point P.

1. 2. 𝑑𝑞=𝜆𝑑𝑥

3.𝑉=∫ 𝑘 𝜆𝑑𝑥𝑥

=𝑘 𝜆 ∫𝑎

𝑎+𝐿

𝑥−1𝑑𝑥

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)• Find the Electric Field

@ point P.

P a

L

4 .𝑉=𝑘 𝜆𝑙𝑛𝑥a

a+L

6.

4. Integrate.

6. Plug in the ranges and simplify.

5 .𝑉=𝑘𝑄𝐿𝑙𝑛𝑥

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)• Find the Electric Field

@ point P.

1. 2. 3.𝐸=∫ 𝑘𝑑𝑞

(𝑎¿¿2+𝑥2)𝑥

√(𝑎2+𝑥2)

¿

1. Equation

2. Don’t 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.

𝐸=𝑘𝑥

(𝑎¿¿ 2+𝑥2)− 3

2∫𝑑𝑞 ¿

x

a P

3. Ring (E)• Find the Electric Field

@ point P.

4 .𝐸=𝑘𝑥

(𝑎¿¿2+𝑥2)− 3

2𝑄 ¿4. Integrate.

6. No range since you will probably know the total charge.

5. No charge density to substitute!

x

a P

3. Ring (V)• Find the Electric

Potential @ point P.

1. 1. Equation2. a & x are constants!

3.𝑉=𝑘

(𝑎¿¿2+𝑥2)∫𝑑𝑞 ¿3. Replace dq. Pull out constants. Set Range.

4.𝐸=𝑘

(𝑎¿¿2+𝑥2)𝑄¿

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)• Find the Electric Field

@ point P.

1.

2. 𝑑𝐴=2𝜋𝑟𝑑𝑟

3.𝐸=∫ 𝑘𝑥𝜎 𝑑𝐴

(𝑟 ¿¿2+𝑥2)32

=∫ 𝑘𝜎 𝑥2𝜋 𝑟𝑑𝑟

(𝑟 ¿¿2+𝑥2)32=𝑘𝜎 𝜋 𝑥∫

0

𝑅2𝑟𝑑𝑟

(𝑟 ¿¿2+𝑥2)32

¿¿ ¿

1. Equation

2. Charge Density. Area because we’re moving out concentric circles.

3. Replace dq. Pull out constants. Set Range.

x

r P

4. Disk (E)• Find the Electric Field

@ point P.

4. Integrate.

6. Plug in the ranges and simplify.

5 .𝐸=−2𝑘𝜎𝜋 𝑥1

(𝑟 2+𝑥2)12 0

R5. Substitute in the

total charge density.

4.𝐸=𝑘𝜎𝜋 𝑥∫0

𝑅𝑑𝑈

𝑈32

=𝑘𝜎𝜋 𝑥𝑈

− 12

−12

=𝑘𝜎𝜋 𝑥𝑈

− 12

−12

=−2𝑘𝜎𝜋 𝑥1

𝑈12

0

R

6 .𝐸=2𝑘𝜎𝜋 [1− 𝑥

(𝑟2+𝑥2)12 ]

x

r P

4. Disk (V)• Find the Electric

Potential @ point P.

3.𝑉=∫ 𝑘𝜎 𝑑𝐴

√𝑟2+𝑥2=∫ 𝑘𝜎 2𝜋𝑟𝑑𝑟

√𝑟 2+𝑥2=𝑘𝜎2𝜋∫

0

𝑅𝑟𝑑𝑟

√𝑟2+𝑥2

1. Equation

2. Charge Density: Area Again

1. 2. 𝑑𝐴=2𝜋𝑟𝑑𝑟

3. Replace dq. Pull out constants. Set Range.

x

r P

4. Disk(V)• Find the Electric Field

@ point P.

4. Integrate.

6. Plug in the ranges and simplify.6 .𝑉=2𝑘𝜎𝜋 (√(𝑟 2+𝑥2)    − x  )

4 .𝑉=𝑘𝜎 2𝜋∫0

𝑅𝑟𝑑𝑟

√𝑟 2+𝑥2=𝑘𝜎 2𝜋∫

0

𝑅 𝑟𝑑𝑈2𝑟

𝑈12

=𝑘𝜎𝜋∫0

𝑅𝑑𝑈

𝑈12

=𝑘𝜎𝜋𝑈

12

12

0

R¿2𝑘𝜎𝜋 √(𝑟2+𝑥2)  

x

r P

5. Arc Length (E)• Find the Electric Field

@ point P.

1.

2. 𝑑𝑞=𝜆𝑑𝑙

3.𝐸=∫ 𝑘𝜆𝑑𝑙𝑟2 𝑐𝑜𝑠𝜃=

𝑘 𝜆𝑟2 ∫

𝜃1

𝜃2

𝑐𝑜𝑠𝜃 𝑑𝑙

1. Equation (cos because all the y values cancel out.)

2. Charge Density.

𝐸=𝑘 𝜆𝑟2 ∫

𝜃1

𝜃2

𝑐𝑜𝑠𝜃𝑟𝑑 𝜃=𝑘 𝜆𝑟 ∫

𝜃 1

𝜃 2

𝑐𝑜𝑠𝜃𝑑𝜃

θ changes with respect to the arc length!

3. Replace dq. Pull out constants (r is constant!). Set Range.

x

rP𝑙

5. Arc Length (E)• Find the Electric Field

@ point P.

6.

4. Integrate.

6. Plug in the ranges and simplify.

5 .𝐸=𝑘𝑄𝐿𝑠𝑖𝑛𝜃

𝜃1

𝜃2 5. Substitute in the total charge density.

𝜃1

𝜃24.𝐸=

𝑘 𝜆𝑟

𝑠𝑖𝑛𝜃

7. 7. The length is the arc length.

x

rP𝑙

5. Arc Length (V)• Find the Electric Field

@ point P.

1. 2. 𝑑𝑞=𝜆𝑑𝑙

3.𝑉=∫ 𝑘 𝜆𝑑𝑙𝑟

=𝑘 𝜆𝑟 ∫

𝜃1

𝜃2

𝑟𝑑𝜃=𝑘 𝜆∫𝜃1

𝜃2

𝑑𝜃

1. Equation

2. Charge Density.

3. Replace dq. Pull out constants (r is constant!). Set Range.θ changes with respect to the arc length! Must be in Radians though (since θ is not in a function.)

x

rP𝑙

5. Arc Length (V)• Find the Electric Field

@ point P.

6.

4. Integrate.

6. Plug in the ranges and simplify.

5 .𝐸=𝑘𝑄𝐿𝜃𝜃1

𝜃2 5. Substitute in the total charge density.

𝜃1

𝜃24.𝑉=𝑘 𝜆𝜃

7. 7. The length is the arc length.

x

rP𝑙