A hollow sphere of radius a carries a total charge Q distributed uniform over its surface, A small area dA of the sphere is cut off. Find the electric field at the center due to the remaining sphere
Answer
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Hint: The space around a charge is always under stress and experiences a force on another charge when placed there. The region or space in which stress exists is called the electric field (or dielectric or electrostatic) field. The stress in the space is represented by the lines of force. Electric field is a vector quantity.
Complete step-by-step solution:
Coulomb’s law states that the magnitude of electrostatic force between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them
$\left| F \right|=k\dfrac{\left| {{q}_{1}}{{q}_{2}} \right|}{{{d}^{2}}}$
Where coulomb’s constant$(k)=\dfrac{1}{4\pi {{\varepsilon }_{0}}}$
Consider a hollow sphere of radius a carries a total charge Q. There will be zero electric field at the center.
dA area is cut meaning that the + charge disappears on dA.
On a small area dA a small amount of negative charge equal to \[\dfrac{QdA}{4\pi {{a}^{2}}}\] is applied this will make dA neutral.
Positive charge does not give any electric field at center,
The negative charge will give an electric field.
Negative charge is at equal distance ‘a’ from center. now apply coulomb’s law
$F=qE$
$E=\dfrac{q}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}$
$E=\dfrac{\dfrac{QdA}{4\pi {{a}^{2}}}}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}$
$E=\dfrac{QdA}{16{{\pi }^{2}}{{\varepsilon }_{0}}{{a}^{4}}}$
The electric field at the center due to the remaining sphere is $\dfrac{QdA}{16{{\pi }^{2}}{{\varepsilon }_{0}}{{a}^{4}}}$
Note: Student’s the magnitude of electric field is obtained by coulomb’s law and in case of a unit stationary point charge coulomb’s law derivation formula is similar to gauss law derivation. When two charged bodies are placed near each other, mechanical force is experienced on them. Force is a vector quantity.
Complete step-by-step solution:
Coulomb’s law states that the magnitude of electrostatic force between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them
$\left| F \right|=k\dfrac{\left| {{q}_{1}}{{q}_{2}} \right|}{{{d}^{2}}}$
Where coulomb’s constant$(k)=\dfrac{1}{4\pi {{\varepsilon }_{0}}}$
Consider a hollow sphere of radius a carries a total charge Q. There will be zero electric field at the center.
dA area is cut meaning that the + charge disappears on dA.
On a small area dA a small amount of negative charge equal to \[\dfrac{QdA}{4\pi {{a}^{2}}}\] is applied this will make dA neutral.
Positive charge does not give any electric field at center,
The negative charge will give an electric field.
Negative charge is at equal distance ‘a’ from center. now apply coulomb’s law
$F=qE$
$E=\dfrac{q}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}$
$E=\dfrac{\dfrac{QdA}{4\pi {{a}^{2}}}}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}$
$E=\dfrac{QdA}{16{{\pi }^{2}}{{\varepsilon }_{0}}{{a}^{4}}}$
The electric field at the center due to the remaining sphere is $\dfrac{QdA}{16{{\pi }^{2}}{{\varepsilon }_{0}}{{a}^{4}}}$
Note: Student’s the magnitude of electric field is obtained by coulomb’s law and in case of a unit stationary point charge coulomb’s law derivation formula is similar to gauss law derivation. When two charged bodies are placed near each other, mechanical force is experienced on them. Force is a vector quantity.
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