91Ó°ÊÓ

The given data is listed below as:

• The Value of the negative point charge is,$q=1×{10}^{-9}C$
• The position of the electric field is,
• The position of charge is,

The electric field is described as a region which helps a charged particle to exert force on another charged particle.

The electric field inside a sphere having uniform charge is expressed as:

$\mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\mathbf{-}\frac{\mathbf{q}}{\left|\stackrel{→}{r}\right|}\hat{\mathbf{r}}$ …(¾±)

Here,$\mathbf{q}$is the charge of the sphere,$\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}$is the electric field constant with value $\mathbf{9}\mathbf{×}{\mathbf{10}}^{\mathbf{9}}\mathbf{N}\mathbf{.}{\mathbf{m}}^{\mathbf{2}}\mathbf{/}\mathbf{C}$,$\left|\stackrel{→}{r}\right|$is the magnitude of the position vector and$\hat{\mathbf{r}}$is the unit vector.

The equation of the value of the position vector is expressed as:

$\stackrel{→}{r}=r_2-Ir$

Here,$r_1$is the location of the charge and $r_2$is the position of the electric field.

Substitute the values in the above equation.

The equation of the magnitude of the position magnitude of vector is expressed as:

$\left|\stackrel{→}{r}\right|=\sqrt{{\left(X_1\right)}^2+{\left(y_1\right)}^2+{\left(Z_1\right)}^2}$

Here,$x_1$is the position of the point in the $x$coordinate, $y_1$is the position of the point in the $y$coordinate and role="math" localid="1656919307064" $z_1$is the position of the point in the z coordinate.

Substitute the values in the above equation.

The equation of the unit vector is expressed as:

$\hat{r}=\frac{\stackrel{→}{r}}{\left|\stackrel{→}{r}\right|}$

Here,$\stackrel{→}{r}$is the position vector and $\left|\stackrel{→}{r}\right|$is the magnitude of the position vector.

Substitute the values in the above equation.

Substitute all the values in the equation (i).

Thus, the magnitude the electric field is .