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Question: Electric field at a distance \(x\) from the origin it is given as \(E = \dfrac{{100\,N - {m^2}{C^{ -...

Electric field at a distance xx from the origin it is given as E=100 N−m2C−1x2E = \dfrac{{100\,N - {m^2}{C^{ - 1}}}}{{{x^2}}}. Then, what is the potential difference between the points situated at x=10 mx = 10\,m and x=20 mx = 20\,m?
(A) 5 V5\,V
(B) 10 V10\,V
(C) 15 V15\,V
(D) 4 V4\,V

Explanation

Solution

Hint
In this problem Electric field value is given and we have to find the potential difference between two points. So, we have to take integration, to find the potential difference between two points. And, by using the relationship between electric field and potential difference, the potential difference between two points can be determined.
Relation between electric field and potential-
E=−dvdxE = \dfrac{{ - dv}}{{dx}}
Where, EE is the electric field, vv is the potential difference, xx is the direction of the charge moved.

Complete step by step solution
Given that, Electric field, E=100x2E = \dfrac{{100}}{{{x^2}}}
Range of potential difference, x=10 mx = 10\,m and x=20 mx = 20\,m
Relation between electric field and potential,
E=−dvdx ....................(1)E = \dfrac{{ - dv}}{{dx}}\,....................\left( 1 \right)
Here, we have to find the potential difference so keep the potential difference in one side and other terms in other side,
−dv=E×dx- dv = E \times dx
The above equation is also written as,
dv=−(E×dx) ...................(2)dv = - \left( {E \times dx} \right)\,...................\left( 2 \right)
By taking integration on both sides,
∫dv=∫−(E×dx)\int {dv = \int { - \left( {E \times dx} \right)} }
The potential difference between x=10 mx = 10\,m to x=20 mx = 20\,m, so apply this value as the limits of the integration, then,
∫2010dv=∫2010−(E×dx) ...................(3)\int\limits_{20}^{10} {dv} = \int\limits_{20}^{10} { - \left( {E \times dx} \right)} \,...................\left( 3 \right)
By taking the negative symbol outside the integration,
∫2010dv=−∫2010(E×dx)\int\limits_{20}^{10} {dv} = - \int\limits_{20}^{10} {\left( {E \times dx} \right)}
Now, substitute the value of EE in the above equation, then the above equation is written as,
∫2010dv=−∫2010100x2dx\int\limits_{20}^{10} {dv} = - \int\limits_{20}^{10} {\dfrac{{100}}{{{x^2}}}dx}
By taking the x2{x^2} from denominator to numerator, then,
∫2010dv=−∫2010100×(x−2)dx\int\limits_{20}^{10} {dv} = - \int\limits_{20}^{10} {100 \times \left( {{x^{ - 2}}} \right)dx}
Now taking the constant value outside the integration,
∫2010dv=−100∫2010x−2dx\int\limits_{20}^{10} {dv} = - 100\int\limits_{20}^{10} {{x^{ - 2}}dx}
Now using integration on both sides, then the above equation is written as,
[v]2010=−100[x−1−1]2010\mathop {\left[ v \right]}\nolimits_{20}^{10} = - 100\mathop {\left[ {\dfrac{{{x^{ - 1}}}}{{ - 1}}} \right]}\nolimits_{20}^{10}
On further simplification,
[v]2010=100[1x]2010\mathop {\left[ v \right]}\nolimits_{20}^{10} = 100\mathop {\left[ {\dfrac{1}{x}} \right]}\nolimits_{20}^{10}
On substituting the limits,
v10−v20=100[110−120]{v_{10}} - {v_{20}} = 100\left[ {\dfrac{1}{{10}} - \dfrac{1}{{20}}} \right]
On simplifying,
v10−v20=100[20−1010×20]{v_{10}} - {v_{20}} = 100\left[ {\dfrac{{20 - 10}}{{10 \times 20}}} \right]
On further calculation,
v10−v20=100[10200]{v_{10}} - {v_{20}} = 100\left[ {\dfrac{{10}}{{200}}} \right]
On multiplying,
v10−v20=1000200{v_{10}} - {v_{20}} = \dfrac{{1000}}{{200}}
By dividing the terms, then the above equation is written as,
v10−v20=5{v_{10}} - {v_{20}} = 5
Thus, the above equation shows the potential difference is 5 V5\,V.
Hence, the option (A) is correct.

Note
While taking the integration, we need to focus because here the power of the term inside the integration is negative. And the substitution of limit must be upper limit and then lower limit. And there must be a subtraction between upper limit and lower limit after the limit substitution.