Question

The figure shows two curves A and B, of which distance. Which of the two represents the graph

one is of the electric field and the other is of the electric potential due to a point charge with

of the electric field?​

Answer 1

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Answer 2

Given:

The figure shows two curves A and B, of which distance.

To find:

Which one denotes electrostatic field intensity and which one denotes electrostatic potential?

Solution:

The general expression for electrostatic field intensity at a distance "r" from a point charge is given as follows (derived from GAUSS' THEOREM):

$\sf{ \therefore \: E = \dfrac{q}{4\pi \epsilon_{0} {r}^{2} } }$

$\boxed{\sf{ = > \: E \propto \dfrac{1}{ {r}^{2} } }}$

Similarly , the general expression for electrostatic field intensity at a distance "r" from a point charge is given as follows :

$\sf{ \therefore \: V = \dfrac{q}{4\pi \epsilon_{0} r} }$

$\boxed{\sf{ = > \: V \propto \dfrac{1}{r} }}$

Now, we can see that both the electrostatic field intensity and electrostatic potential curves will be hyperbolic in nature but:

• Since E is inversely proportional to r² , its curve will be more steep as compared to potential curve.

So, curve B will be E vs r and curve A will be V vs r

HAVE A LOOK AT sample y = 1/x and 1/x² curves. You will understand the explanation.