
What is the charge density on the surface of the conducting sphere of radius 0.15m whose potential is 200V (with V = 0 at infinity).
A. $3.3 \times {10^{ - 9}}\dfrac{C}{{{m^2}}}$
B. $1.2 \times {10^{ - 8}}\dfrac{C}{{{m^2}}}$
C. $3.3 \times {10^{ - 8}}\dfrac{C}{{{m^2}}}$
D. $1.2 \times {10^{ - 9}}\dfrac{C}{{{m^2}}}$
Answer
508.8k+ views
Hint: charge density is defined as the charge present in the unit area. It can be calculated as Charge density = $\dfrac{{ch\arg e}}{{area}} = \dfrac{Q}{A}$. We can calculate the charge with the help of electric potential (V) i.e., V = $\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{R}$.
Complete answer:
Now, from the question
Given a conducting sphere of radius 0.15m;
Potential on the surface = 200V,
We have to find the charge density on the surface of the sphere.
We know that potential on surface of sphere is V = $\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{R}$
Where, Q and R are the charge on the surface of sphere and radius respectively.
So, given V = 200V
200 = $\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{R}$
$Q = 200 \times 4\pi {\varepsilon _0} \times R$
= $\dfrac{{200 \times 0.15}}{{9 \times {{10}^9}}} = 3.33 \times {10^{ - 9}}C$
Charge density is equal to (Q) per unit area of sphere.
= $\dfrac{{3.33 \times {{10}^{ - 9}}}}{{4 \times 3.14 \times {{(0.15)}^2}}}$
= $11.795 \times {10^{ - 9}}\dfrac{C}{{{m^2}}}$
= $1.2 \times {10^{ - 8}}\dfrac{C}{{{m^2}}}$
So, the correct answer is “Option B”.
Note:
According to electromagnetism, charge density is defined as a measure of electrical charge per unit volume of the space in one, two or three dimensions. To be specific, the linear surface or volume charge density is the amount of electrical charge per area or volume, respectively.
Surface charge describes the electrical potential drop between the inner and outer surface of various states like solid and liquid, liquid and gas or gas and liquid. The surface charge density is present only in conducting surfaces and describes the entire amount of charge q per unit area A.
Hence, the Surface charge density formula is given by,
$\sigma = \dfrac{Q}{A}$
Where,
$\sigma $ = surface charge density
q = charge
A = area
Complete answer:
Now, from the question
Given a conducting sphere of radius 0.15m;
Potential on the surface = 200V,
We have to find the charge density on the surface of the sphere.
We know that potential on surface of sphere is V = $\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{R}$
Where, Q and R are the charge on the surface of sphere and radius respectively.
So, given V = 200V
200 = $\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{R}$
$Q = 200 \times 4\pi {\varepsilon _0} \times R$
= $\dfrac{{200 \times 0.15}}{{9 \times {{10}^9}}} = 3.33 \times {10^{ - 9}}C$
Charge density is equal to (Q) per unit area of sphere.
= $\dfrac{{3.33 \times {{10}^{ - 9}}}}{{4 \times 3.14 \times {{(0.15)}^2}}}$
= $11.795 \times {10^{ - 9}}\dfrac{C}{{{m^2}}}$
= $1.2 \times {10^{ - 8}}\dfrac{C}{{{m^2}}}$
So, the correct answer is “Option B”.
Note:
According to electromagnetism, charge density is defined as a measure of electrical charge per unit volume of the space in one, two or three dimensions. To be specific, the linear surface or volume charge density is the amount of electrical charge per area or volume, respectively.
Surface charge describes the electrical potential drop between the inner and outer surface of various states like solid and liquid, liquid and gas or gas and liquid. The surface charge density is present only in conducting surfaces and describes the entire amount of charge q per unit area A.
Hence, the Surface charge density formula is given by,
$\sigma = \dfrac{Q}{A}$
Where,
$\sigma $ = surface charge density
q = charge
A = area
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