
For the case of a charged non-conducting sphere, what plot represents correctly the distribution of electric field E
Answer
572.1k+ views
Hint:The above problem can be resolved using the concepts and the fundamentals of the electric field intensity for a solid cylinder. The magnitude of the intensity of electric field changes, while moving from the centre of the sphere to any of the points outside the sphere. Moreover, the magnitude becomes exactly null, when any point at the centre is considered.
Complete step by step answer:
Consider a charged non-conducting sphere of radius a. For which the expression of electric field for the different cases are mentioned as:
Case - 1
The expression for the magnitude of electric field at a point outside the solid sphere is,
\[E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}\]
Here, r is the distance of point outside the sphere from the centre of the sphere.
Case – 2
The electric field at a point on surface of solid sphere is,
\[E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{a^2}}}\]
Here, a is the point on the sphere.
Case -3
The electric field at a point inside the solid sphere is,
\[E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{qr}}{{{a^3}}}\]
Here, a is the point inside the sphere.
Case-4
At the centre of the solid sphere, the magnitude of the electric field is zero. That is,
\[E = 0\]
Therefore, with the reference of above all the condition, the graph plotted in option (B) is correct.
Note:Try to understand the concept of electric field and its significance in the case of a solid sphere. To determine the intensity of electric field the region is analysed, where the magnitude of the electrostatic force is experienced within the charged species. The electric field has its application in various places in physics that vary accordingly with the electrostatic force.
Complete step by step answer:
Consider a charged non-conducting sphere of radius a. For which the expression of electric field for the different cases are mentioned as:
Case - 1
The expression for the magnitude of electric field at a point outside the solid sphere is,
\[E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}\]
Here, r is the distance of point outside the sphere from the centre of the sphere.
Case – 2
The electric field at a point on surface of solid sphere is,
\[E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{a^2}}}\]
Here, a is the point on the sphere.
Case -3
The electric field at a point inside the solid sphere is,
\[E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{qr}}{{{a^3}}}\]
Here, a is the point inside the sphere.
Case-4
At the centre of the solid sphere, the magnitude of the electric field is zero. That is,
\[E = 0\]
Therefore, with the reference of above all the condition, the graph plotted in option (B) is correct.
Note:Try to understand the concept of electric field and its significance in the case of a solid sphere. To determine the intensity of electric field the region is analysed, where the magnitude of the electrostatic force is experienced within the charged species. The electric field has its application in various places in physics that vary accordingly with the electrostatic force.
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