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Recall that:
[tex]\Delta V = -\int\limits^a_b {E \cdot } \, dx[/tex]
Given an electric field, the potential difference can be solved by using integration. Similarly:
[tex]E = -\frac{dV}{dx}[/tex]
We can differentiate the electric potential equation to solve for the electric field.
Use the power rule:
[tex]\frac{dy}{dx} x^n = nx^{n - 1}[/tex]
Differentiate the given equation.
[tex]-\frac{dV}{dx}\frac{7}{x^2} =- \frac{dV}{dx}7x^{-2} = -(-14x^{-3}) = \frac{14}{x^3}[/tex]
Or:
[tex]\boxed{E(x) = \frac{14}{x^3}}[/tex]
The component of the electric field in that region is E = 14/x³ if the electric potential in a region is given by v(x)=7/x2
An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.
We have:
The electric potential in a region is given by:
[tex]\rm V(x) = \frac{7}{x^2}[/tex]
We know the relation between potential difference and an electric field is given by:
[tex]\rm \triangle V = \int\limits^a_b {E.} \, dx[/tex]
If the electric field is given, then:
[tex]\rm E = -\frac{dV}{dx}[/tex]
[tex]\rm E = -\frac{d(7x^{-2} )}{dx}[/tex]
After differentiate:
[tex]\rm E = -(-14x^{-3})[/tex] or
[tex]\rm E = \frac{14}{x^3}[/tex]
Thus, the component of the electric field in that region is E = 14/x³ if the electric potential in a region is given by v(x)=7/x2
Learn more about the electric field here:
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