The dielectric strength of air, E = 3.0×106 V/m, is the maximum field that air can withstand before it breaks down and becomes conducting.
Part A) How much charge can be placed on a spherical conductor with a 15.0- cm radius before the field at its surface exceeds the breakdown strength of the air?
Part B) What would be the electric potential at the surface of this conductor?
The Correct Answer and Explanation is :
To solve this problem, we will use the following equations from electrostatics:
- Electric field at the surface of a spherical conductor:
[
E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}
]
where:
- ( E = 3.0 \times 10^6 ) V/m (dielectric strength of air),
- ( Q ) is the charge on the conductor,
- ( r = 15.0 ) cm = ( 0.15 ) m (radius of the sphere),
- ( \epsilon_0 = 8.85 \times 10^{-12} ) C²/N·m² (permittivity of free space).
- Electric potential at the surface of a spherical conductor:
[
V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}
]
Part A: Maximum Charge Before Breakdown
Rearranging the electric field equation for ( Q ):
[
Q = E \cdot 4\pi\epsilon_0 \cdot r^2
]
Substituting the values:
[
Q = (3.0 \times 10^6) \cdot (4\pi \cdot 8.85 \times 10^{-12}) \cdot (0.15)^2
]
[
Q = (3.0 \times 10^6) \cdot (1.112 \times 10^{-10}) \cdot (0.0225)
]
[
Q = 7.5 \times 10^{-6} \text{ C} = 7.5 \mu C
]
Part B: Electric Potential at the Surface
Using the potential equation:
[
V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}
]
Substituting the values:
[
V = \frac{(7.5 \times 10^{-6})}{(4\pi \times 8.85 \times 10^{-12} \times 0.15)}
]
[
V = \frac{(7.5 \times 10^{-6})}{(1.67 \times 10^{-12})}
]
[
V = 4.49 \times 10^6 \text{ V} = 4.49 \text{ MV}
]
Explanation (300 words)
A conductor can hold charge until the surrounding medium (in this case, air) experiences electrical breakdown, meaning that the air becomes ionized and starts conducting electricity. This breakdown occurs when the electric field at the conductor’s surface exceeds the dielectric strength of air, which is ( 3.0 \times 10^6 ) V/m.
To find the maximum charge that can be placed on a spherical conductor of radius ( 15.0 ) cm, we use the equation for the electric field just outside a charged sphere. The charge ( Q ) produces a radial electric field that follows an inverse-square law. Rearranging the equation to solve for ( Q ), we find that the maximum charge is ( 7.5 ) microcoulombs.
Once we know the maximum charge, we calculate the electric potential at the surface. The potential of a charged sphere is directly proportional to the charge and inversely proportional to its radius. With the obtained charge value, the potential at the surface of the sphere is found to be approximately ( 4.49 ) million volts (MV).
This calculation is important in designing high-voltage equipment like Van de Graaff generators or lightning rods, where charge buildup must be controlled to prevent unintended electrical discharge. If the charge on a conductor exceeds this limit, it will ionize the surrounding air, leading to a spark or electrical breakdown, which is why insulators or controlled environments are used to handle high-voltage situations.