Respuesta :
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
[tex]F = qE[/tex]....(I)
Using formula of electric force
[tex]F =\dfrac{kq^2}{d^2}[/tex]....(II)
From equation (I) and (II)
[tex]qE=\dfrac{kq^2}{d^2}[/tex]
[tex]E=\dfrac{kq}{d^2}[/tex]
[tex]q=\dfrac{E(x-r)^2}{k}[/tex]...(III)
For both points,
[tex]\dfrac{E(x-r)^2}{k}=\dfrac{E(x-r)^2}{k}[/tex]
Put the value into the formula
[tex]\dfrac{10.0\times(x-5.00)^2}{k}=\dfrac{17.0\times(x-10.0)^2}{k}[/tex]
[tex]10.0\times(x-5.0)^2=17.0\times(x-10.0)^2[/tex]
Take the square root of both sides
[tex]3.162(x-5.0)=4.123(x-10.0)[/tex]
[tex]3.162x-3.162\times5.0=4.123x-4.123\times10.0[/tex]
[tex]3.162x-4.123x=-4.123\times10.0+3.162\times5.0[/tex]
[tex]0.961x=25.42[/tex]
[tex]x=\dfrac{25.42}{0.961}[/tex]
[tex]x=26.45\ cm[/tex]
(B). We need to calculate the charge
Using equation (III)
[tex]q=\dfrac{E(x-r)^2}{k}[/tex]
Put the value into the formula
[tex]q=\dfrac{10.0(26.45\times10^{-2}-5.00\times10^{-2})^2}{9\times10^{9}}[/tex]
[tex]q=5.11\times10^{-11}\ C[/tex]
[tex]q=51.1\ pC[/tex]
Hence, (A). The location of the charge is 26.45 cm.
(B). The magnitude of the charge is 51.1 pC.
