Question

The electric field at the point x=5.00cm and y=0 points in the positive x direction with a magnitude of 10.0 N/C . At the point x=10.0cm and y=0 the electric field points in the positive x direction with a magnitude of 17.0 N/C . Assume this electric field is produced by a single point charge. Part AFind the charge's location.Part BFind the magnitude of the charge.

Answer 1

Answer:

(A). The location of the charge is 26.45 cm.

(B). The magnitude of the charge is 51.1 pC.

Explanation:

Given that,

Distance in x axis = 5.00 cm

Electric field = 10.0 N/C

Distance in x axis = 10.0 cm

Electric field = 17.0 N/C

Since, q is the same charge, two formulas can be set equal using the two different electric fields.

(a). We need to calculate the location of the charge

Using formula of force

$F = qE$....(I)

Using formula of electric force

$F =\dfrac{kq^2}{d^2}$....(II)

From equation (I) and (II)

$qE=\dfrac{kq^2}{d^2}$

$E=\dfrac{kq}{d^2}$

$q=\dfrac{E(x-r)^2}{k}$...(III)

For both points,

$\dfrac{E(x-r)^2}{k}=\dfrac{E(x-r)^2}{k}$

Put the value into the formula

$\dfrac{10.0\times(x-5.00)^2}{k}=\dfrac{17.0\times(x-10.0)^2}{k}$

$10.0\times(x-5.0)^2=17.0\times(x-10.0)^2$

Take the square root of both sides

$3.162(x-5.0)=4.123(x-10.0)$

$3.162x-3.162\times5.0=4.123x-4.123\times10.0$

$3.162x-4.123x=-4.123\times10.0+3.162\times5.0$

$0.961x=25.42$

$x=\dfrac{25.42}{0.961}$

$x=26.45\ cm$

(B). We need to calculate the charge

Using equation (III)

$q=\dfrac{E(x-r)^2}{k}$

Put the value into the formula

$q=\dfrac{10.0(26.45\times10^{-2}-5.00\times10^{-2})^2}{9\times10^{9}}$

$q=5.11\times10^{-11}\ C$

$q=51.1\ pC$

Hence, (A). The location of the charge is 26.45 cm.

(B). The magnitude of the charge is 51.1 pC.