Answer:
Surface charge density = 1.43 × 10⁶μC/m²
Explanation:
surface charge density = Q /A_____(1)
where charge Q is the uniformly distributed on surface area A and d surface charge density σ
The electric field due to uniformly charge sphere of charge Q a distance r from the center of the sphere
[tex]E = k\frac{Q}{r^2}[/tex]______(2)
where k is 8.988 × 10⁹N.m²C²
The surface area of the sphere is 4πR² ______(3)
The Capacitance is 4πε₀R
where ε₀ = 8.8542 × 10⁻¹²C/Nm² is a constant
Given that,
R = 11cm = 0.11m
E = 4.90 ✕ 10⁴ N/C
r = 20.0cm = 0.20m
substitute for Q in eqn(2) and for A in eqn(3)
surface charge density = [tex]\frac{Er^2 }{k(4\pi R^2)} \\\frac{(4.9 * 10^4)(0.20)^2}{4\pi (8.988 * 10^9)(0.11)^2 }[/tex]
= 1.43 * 10⁶C/m²
Surface charge density = 1.43 μC/m²
An isolated, charged conducting sphere of radius 11.0 cm creates an electric field of 4.90×10⁴ N/C at a distance 20.0 cm from its center. (a) What is its surface charge density? (b) What is its capacitance?
(a) 1.47 x 10⁻⁶ C/m² or 1.47 μC/m²
(b) 12.07 x 10⁻¹² F or 12.07 pF
The surface charge density, σ, of a surface (sphere, in this case) of area A which has a charge Q uniformly distributed on it is given by;
σ = [tex]\frac{Q}{A}[/tex] -----------------(i)
Also, the electric field, E, due to the charge Q, at a distance r, from the center of the sphere to another point on the sphere is given as;
E = k x [tex]\frac{Q}{r^2}[/tex] --------------(ii)
Where;
k = Coulomb's constant = 8.99 x 10⁹Nm²/C²
(a) i. First calculate the charge on the sphere as follows;
From the question;
r = 20.0cm = 0.20m
E = 4.90 x 10⁴ N/C
Substitute these values into equation (ii) as follows;
4.90 x 10⁴ = 8.99 x 10⁹ x [tex]\frac{Q}{0.20^2}[/tex]
4.90 x 10⁴ = 8.99 x 10⁹ x [tex]\frac{Q}{0.04}[/tex]
4.90 x 10⁴ = 224.75 x 10⁹ x Q
Q = [tex]\frac{4.90*10^4}{224.75*10^9}[/tex]
Q = 0.022 x 10⁻⁵
Q = 0.22 x 10⁻⁶ C
(a) ii. Also calculate the area A, of the sphere as follows;
A = 4π R²
Where;
R = radius of the sphere = 11.0cm = 0.11m
Substitute this value into equation above;
A = 4π (0.11)² [Take π = 3.142]
A = 4(3.142)(0.0121)
A = 0.15m²
(a) iii. Now calculate the surface charge density, of the sphere as follows;
Substitute the values of A and Q into equation (i) as follows;
σ = [tex]\frac{0.22 * 10^{-6}}{0.15}[/tex]
σ = 1.47 x 10⁻⁶C/m²
Therefore the surface charge density is 1.47 x 10⁻⁶C/m²
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(b) The capacitance C, of an isolated charged sphere with radius R, is given by;
C = Aε₀ / R ----------------(iii)
Where;
R = 11.0cm = 0.11m
A = area of the sphere = 0.15m² [as calculated above]
ε₀ = permittivity of free space = 8.85 x 10⁻¹² C/Nm² [a known constant]
Substitute these values into equation (iii) as follows;
C = 0.15 x 8.85 x 10⁻¹² / 0.11
C = 12.07 x 10⁻¹²F
Therefore, the capacitance of the charged sphere is 12.07 x 10⁻¹²F
