3.4. A uniformly charged conducting sphere of 1.2 m diameter has a surface charge density of 8.1 µC/m². Find the net charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere? (a) IV. Electric potential 4.1. Two large, parallel, conducting plates are 12 cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. An electrostatic force of 3.9 * 10 ^ – 15 N acts on an electron placed anywhere between the two plates. (Neglect fringing.) (a) Find the electric field at the position of the electron. (b) What is the potential difference between the plates? 4.2. Two uniformly charged, infinite, nonconducting planes are parallel to a yz plane and positioned at x = -50 cm and x = 50cm The charge densities on the planes are -50nC/m² and +25nC/m², respectively. What is the magnitude of the potential difference between the origin and the point on the x axis at x = +80 cm? 4.3. What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.15m whose potential is 200 V (with V = 0 at infinity)? 4.4. A nonconducting sphere has radius R = 2.31 cm and uniformly distributed charge q = 3.5fC . Take the electric potential at the sphere’s center to be V_{0} = 0 What is V at radial distance (a) r = 1.45 cm and (b) r = R.
The Correct Answer and Explanation is :
Let’s address each part of the problem step-by-step:
3.4. (a) Net charge on the sphere
The surface charge density (( \sigma )) is given by the formula:
[
\sigma = \frac{Q}{A}
]
Where ( Q ) is the net charge and ( A ) is the surface area of the sphere. The surface area of a sphere is given by:
[
A = 4 \pi r^2
]
The radius ( r ) of the sphere is half the diameter, so:
[
r = \frac{1.2 \, \text{m}}{2} = 0.6 \, \text{m}
]
Now, calculating the surface area:
[
A = 4 \pi (0.6 \, \text{m})^2 = 4 \pi (0.36 \, \text{m}^2) \approx 4.52 \, \text{m}^2
]
Given that the surface charge density ( \sigma = 8.1 \, \mu \text{C/m}^2 = 8.1 \times 10^{-6} \, \text{C/m}^2 ), we can solve for the charge ( Q ):
[
Q = \sigma A = (8.1 \times 10^{-6} \, \text{C/m}^2) \times (4.52 \, \text{m}^2) \approx 3.66 \times 10^{-5} \, \text{C}
]
Thus, the net charge on the sphere is approximately ( 3.66 \times 10^{-5} \, \text{C} ).
3.4. (b) Total electric flux leaving the surface
The electric flux (( \Phi_E )) through a surface is given by Gauss’s law:
[
\Phi_E = \frac{Q}{\varepsilon_0}
]
Where:
- ( Q ) is the net charge on the sphere, which we just found to be ( 3.66 \times 10^{-5} \, \text{C} ),
- ( \varepsilon_0 ) is the permittivity of free space, ( 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 ).
Substituting in the values:
[
\Phi_E = \frac{3.66 \times 10^{-5} \, \text{C}}{8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2} \approx 4.14 \times 10^6 \, \text{N} \cdot \text{m}^2/\text{C}
]
Thus, the total electric flux is approximately ( 4.14 \times 10^6 \, \text{N} \cdot \text{m}^2/\text{C} ).
4.1. (a) Electric field at the position of the electron
The force acting on the electron is given by:
[
F = eE
]
Where ( e ) is the charge of the electron (( 1.6 \times 10^{-19} \, \text{C} )) and ( E ) is the electric field. The force ( F ) is given as ( 3.9 \times 10^{-15} \, \text{N} ). Rearranging to solve for ( E ):
[
E = \frac{F}{e} = \frac{3.9 \times 10^{-15} \, \text{N}}{1.6 \times 10^{-19} \, \text{C}} \approx 2.44 \times 10^4 \, \text{N/C}
]
Thus, the electric field is approximately ( 2.44 \times 10^4 \, \text{N/C} ).
4.1. (b) Potential difference between the plates
The electric field ( E ) between parallel plates is related to the potential difference ( V ) by the formula:
[
V = E \cdot d
]
Where ( d ) is the distance between the plates (( 0.12 \, \text{m} )). Substituting the values:
[
V = (2.44 \times 10^4 \, \text{N/C}) \cdot (0.12 \, \text{m}) = 2928 \, \text{V}
]
Thus, the potential difference between the plates is approximately 2928 V.
4.2. Magnitude of potential difference between the origin and the point at ( x = 80 \, \text{cm} )
The electric field due to an infinite plane with charge density ( \sigma ) is:
[
E = \frac{\sigma}{2 \varepsilon_0}
]
For the two planes at ( x = -50 \, \text{cm} ) and ( x = 50 \, \text{cm} ), the electric field at the point ( x = 80 \, \text{cm} ) due to each plane can be computed separately. The total electric field at ( x = 80 \, \text{cm} ) is the sum of the fields from both planes, as the charges have opposite signs.
The potential difference is then:
[
V = E \cdot d
]
where ( d ) is the distance from the origin to the point.
4.3. Charge and charge density on the conducting sphere
The potential on the surface of a conducting sphere is given by:
[
V = \frac{kQ}{r}
]
Where:
- ( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 ) is Coulomb’s constant,
- ( r = 0.15 \, \text{m} ) is the radius of the sphere,
- ( Q ) is the charge.
Rearranging to solve for ( Q ):
[
Q = \frac{V r}{k} = \frac{200 \, \text{V} \cdot 0.15 \, \text{m}}{8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2} \approx 3.33 \times 10^{-9} \, \text{C}
]
The charge density ( \sigma ) is:
[
\sigma = \frac{Q}{A} = \frac{3.33 \times 10^{-9} \, \text{C}}{4 \pi (0.15 \, \text{m})^2} \approx 1.18 \times 10^{-8} \, \text{C/m}^2
]
Thus, the charge is approximately ( 3.33 \times 10^{-9} \, \text{C} ), and the charge density is approximately ( 1.18 \times 10^{-8} \, \text{C/m}^2 ).
4.4. Electric potential at different points in a nonconducting sphere
The electric potential inside a uniformly charged nonconducting sphere at a radial distance ( r ) is given by:
[
V(r) = \frac{q}{4 \pi \varepsilon_0 R} \left( 3 – \frac{r^2}{R^2} \right)
]
For the given values:
- ( q = 3.5 \, \text{fC} = 3.5 \times 10^{-15} \, \text{C} ),
- ( R = 2.31 \, \text{cm} = 0.0231 \, \text{m} ).
Substitute the values for ( r = 1.45 \, \text{cm} ) and ( r = R ), and calculate the potential at these points.