Question

At a distance r from a point located at origin

in space, the electric potential varies as V = 10r.

Find the electric field at = 3+ 4− 5.​

Answer 1

The electric field is $\boxed{-\sqrt{2}(3\hat i+4\hat j-5\hat k)}$

Explanation:

Given

The electric potential varies as $V=10r$

To find out

Electric potential at $3\hat i+4\hat j-5\hat k$

Solution:

$V=10r$

$\vec r=x\hat i+y\hat j+z\hat k$

$\implies r=|\vec r|=\sqrt{x^2+y^2+z^2}$

Therefore,

$V=10\sqrt{x^2+y^2+z^2}$

Electric field $\vec E$ is given as

$\vec E=E_x\hat i+E_y\hat j+E_z\hat k$

Where

$E_x=-\frac{dV}{dx}=-10\frac{d}{dx}(\sqrt{x^2+y^2+z^2})=-10\frac{2x}{2\sqrt{x^2+y^2+z^2}}$

$E_x=-\frac{dV}{dy}=-10\frac{d}{dy}(\sqrt{x^2+y^2+z^2})=-10\frac{2y}{2\sqrt{x^2+y^2+z^2}}$

$E_x=-\frac{dV}{dx}=-10\frac{d}{dz}(\sqrt{x^2+y^2+z^2})=-10\frac{2z}{2\sqrt{x^2+y^2+z^2}}$

Thus,

$\vec E=-10\frac{(x\hat i+y\hat j+z\hat k)}{\sqrt{x^2+y^2+z^2}}$

At the given point $3\hat i+4\hat j-5\hat k$, the electric potential will be

$\vec E=-10\times \frac{(3\hat i+4\hat j-5\hat k)}{\sqrt{3^2+4^2+(-5)^2}}$

$\implies \vec E=-10\times \frac{(3\hat i+4\hat j-5\hat k)}{5\sqrt{2}}$

$\implies \vec E=-10\sqrt{2}\times \frac{(3\hat i+4\hat j-5\hat k)}{10}$

$\implies \vec E=-\sqrt{2}(3\hat i+4\hat j-5\hat k)$

Hope this answer is helpful.

Know More:

Q: An electric field of 20n/c exist along the direction of 3i-4j+5k in space . calculate potential difference between.

Click Here: https://brainly.in/question/3728666

Answer 2

in attachment.

Explanation:

Thank you:-)