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Grade 12th PassElectric Current

The electric potential at a point (x,y,z) is given by V = a + bx^n volt (a and b are positive constants). The magnitude of electric field at (1,2,3) is 2 N / C and at (2,3,4) is 8 N / C. The electric field at (3,4,5) is? Pls answer with a brief explanation....

Sridev Venkataraman

7 Years agoGrade 12th Pass

1 Answer

Samyak Jain

7 Years ago

We know that electric field intensity at a point (x,y,z) is given by E = – [( $\delta$ V/ $\delta$ x)i + ( $\delta$ V/ $\delta$ y)j + ( $\delta$ V/ $\delta$ z)k].

Hence, magnitude of electric field at any point is given by |dV/dr| in r direction.

Here, electric potential varies in x direction only, so we shall substitute dr as dx. Otherwise we need to put dr

as $\delta$ x, $\delta$ y, $\delta$ z separately to get electric field vector.

Now, V = a + b xⁿ $\Rightarrow$ dV/dx = bn x^{n – 1}.

|dV/dx| at (1,2,3) is bn(1)^{n – 1} = 2 N/C $\Rightarrow$ bn = 2 ...(1)

|dV/dx| at (2,3,4) is bn(2)^{n – 1} = 8 N/C ...(2)

Dividing (2) by (1), we get (2)^{n – 1} = 4 = 2²

$\therefore$ n – 1 = 2 or n = 3. Substitute the value of n in (1) to get b = 2/3.

We get V = a + (2/3) x³

So, |dV/dx| at (x,y,z) is 2 x².

Thus, magnitude of electric field at (3,4,5) is 2.(3)² = 18.

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