A conductor with an inner cavity, shown in Fig. 5, carries a total charge of +8.4𝜇𝐶. Within the cavity, insulated from the conductor, is a point charge of −6 𝜇𝐶.
How much charge is on
4.4.1
the inner surface of the conductor
(3)
4.4.2
the outer surface of the conductor.
For a conductor in electrostatic equilibrium, all excess charge resides on its outer surface.
A conductor with an inner cavity, shown in Fig. 5, carries a total charge of +8.4𝜇𝐶. Within the cavity, insulated from the conductor, is a point charge of −6 𝜇𝐶.How much charge is on
the inner surface of the conductor and
the outer surface of the conductor.
Since the total charge on the conductor is +8.4 µC and the charge inside the cavity is -6 µC, the remaining charge on the inner surface of the conductor is 8.4 µC - (-6 µC) = 14.4 µC.
Since all the excess charge resides on the outer surface of the conductor, the charge on the outer surface of the conductor is also 14.4 µC.
A conductor with an inner cavity, carries a total charge of +8.4𝜇𝐶. Within the cavity, insulated from the conductor, is a point charge of −6 𝜇𝐶.How much charge is on
the inner surface of the conductor and
the outer surface of the conductor.
To find the charge on the inner surface of the conductor, we need to consider the principle of charge conservation. Since the conductor is in electrostatic equilibrium, the total charge on the conductor must equal the sum of the charges on its inner and outer surfaces.
Given that the total charge on the conductor is +8.4 μC and the charge inside the cavity is -6 μC, we can calculate the charge on the inner surface by subtracting the charge inside the cavity from the total charge on the conductor.
Charge on the inner surface = Total charge on the conductor - Charge inside the cavity
= +8.4 μC - (-6 μC)
= +8.4 μC + 6 μC
= +14.4 μC
Therefore, the charge on the inner surface of the conductor is +14.4 μC.
Since the cavity is insulated from the conductor, the charge inside the cavity does not affect the charge distribution on the outer surface of the conductor. Hence, the charge on the outer surface of the conductor is the same as the total charge on the conductor.
Therefore, the charge on the outer surface of the conductor is +8.4 μC.
Electric charge Q is distributed uniformly around a thin ring of radius a as shown in Fig.6. A point P lies on the ring axis at a distance x from the center of the ring.
Determine the electric potential at a point P.
To determine the electric potential at point P, we can break down the ring into infinitesimally small segments and calculate the contribution to the potential at point P due to each segment.
The electric potential at point P due to a small charge element dq on the ring can be calculated using the equation:
dV = k * dq / r
where dV is the potential due to the small charge element dq, k is the electrostatic constant (9 * 10^9 Nm²/C²), and r is the distance from the charge element dq to point P.
Since the charge is uniformly distributed around the ring, the magnitude of charge dq can be expressed as dq = Q * dl / (2πa), where Q is the total charge on the ring and dl is the differential length of the charge element.
Let's assume that the angle between the segment and the line connecting the center of the ring to point P is θ. Then, the differential length dl can be expressed as dl = a * dθ.
Substituting the expressions for dq and dl into the equation for dV, we get:
dV = k * (Q * dl / (2πa)) / r
= k * Q * a * dθ / (2πa * r)
= (k * Q * dθ) / (2πr)
To find the total potential at point P, we need to integrate this equation over the entire ring:
V = ∫(k * Q * dθ) / (2πr) from 0 to 2π
Simplifying the integral:
V = (k * Q) / (2πr) * ∫dθ from 0 to 2π
= (k * Q) / (2πr) * 2π
= k * Q / r
Therefore, the electric potential at point P due to a uniformly charged ring is given by the equation V = k * Q / r, where k is the electrostatic constant, Q is the total charge on the ring, and r is the distance from the center of the ring to point P.
To determine the electric potential at point P, we can break down the ring into infinitesimally small segments and calculate the contribution to the potential at point P due to each segment.
The electric potential at point P due to a small charge element dq on the ring can be calculated using the equation:
dV = k * dq / r
where dV is the potential due to the small charge element dq, k is the electrostatic constant (9 * 10^9 Nm²/C²), and r is the distance from the charge element dq to point P.
Since the charge is uniformly distributed around the ring, the magnitude of charge dq can be expressed as dq = Q * dl / (2πa), where Q is the total charge on the ring and dl is the differential length of the charge element.
Let's assume that the angle between the segment and the line connecting the center of the ring to point P is θ. Then, the differential length dl can be expressed as dl = a * dθ.
Substituting the expressions for dq and dl into the equation for dV, we get:
dV = k * (Q * dl / (2πa)) / r
= k * Q * a * dθ / (2πa * r)
= (k * Q * dθ) / (2πr)
To find the total potential at point P, we need to integrate this equation over the entire ring:
V = ∫(k * Q * dθ) / (2πr) from 0 to 2π
Simplifying the integral:
V = (k * Q) / (2πr) * ∫dθ from 0 to 2π
= (k * Q) / (2πr) * 2π
= k * Q / r
Therefore, the electric potential at point P due to a uniformly charged ring is given by the equation V = k * Q / r, where k is the electrostatic constant, Q is the total charge on the ring, and r is the distance from the center of the ring to point P.Hence, find the electric field at point P.
To find the electric field at point P, we can make use of the electric potential we calculated earlier and use the relationship between electric field and electric potential.
The electric field can be determined by taking the negative gradient of the electric potential, given by:
E = -∇V
In the case of a point charge, the electric field is given by:
E = k * Q / r^2
However, in this case, we are dealing with a uniformly charged ring. To find the electric field at point P due to the ring, we need to take the derivative of the electric potential equation we derived earlier with respect to r:
dV/dr = -k * Q / r^2
Therefore, the electric field at point P due to the uniformly charged ring is given by:
E = -dV/dr = k * Q / r^2
where k is the electrostatic constant, Q is the total charge on the ring, and r is the distance from the center of the ring to point P.
Hence, the electric field at point P is E = k * Q / r^2.
To determine the charges on the inner and outer surfaces of the conductor, we need to understand the concept of electrostatic equilibrium.
In an electrostatically equilibrium conductor, the charge resides only on the outer surface (assuming it is a conductor with no excess charges outside it). This phenomenon is known as the "Faraday's Cage" effect.
So, the charge inside the conductor will redistribute itself in such a way that the electric field inside becomes zero.
Given that the total charge on the conductor is +8.4 𝜇C and there is a point charge of -6 𝜇C within the cavity.
4.4.1:
To find the charge on the inner surface of the conductor, we need to consider that the charge inside the conductor will redistribute itself to neutralize the electric field inside. Since the inner surface of the conductor is closer to the point charge, it will experience the effect most directly.
Therefore, the charge on the inner surface of the conductor will be equal in magnitude and opposite in sign to the point charge within the cavity, which is -6 𝜇C.
Thus, the charge on the inner surface of the conductor is -6 𝜇C.
4.4.2:
The charge on the outer surface of the conductor can be found by subtracting the charge on the inner surface from the total charge on the conductor.
Outer surface charge = Total charge - Inner surface charge
= +8.4 𝜇C - (-6 𝜇C)
= +8.4 𝜇C + 6 𝜇C
= +14.4 𝜇C
Therefore, the charge on the outer surface of the conductor is +14.4 𝜇C.