Question

Three charges are located along y-axis with first -77 C charge at origin, second -100 C charge 2m above origin and third +50 C charge 3m below origin.

(a) What is the magnitude and direction of force exerted by second charge on first charge,

(b) what is the magnitude and direction of force exerted by third charge on first charge,

(c) What is the net force on first charge?

(d) If the third charge is replaced by an imaginary test charge find the magnitude and direction of electric field at its location due to first and second charges

Answer 1

Solution:

Let us name the charges as follows

91 = -77 с.

$q_{2}=-100\; \mathrm{C}$

93 = 50 C

The pictorial representation of the given problem is shown below

92 = -100 C r1 = 2 m 41 = -77 C r2 = 3 m 43 = 50 C

The force exerted by the charge $q_{2}$ on $q_{1}$ can be calculated as follows

$F_{21}=\frac{1}{4\pi \varepsilon _{o}}\frac{q_{1}q_{2}}{r_{1}^{2}}$

Now, substitute the known values

F21 = (9 x 109)_-77 x –100) 22

${\color{Blue} F_{21}=1.7325\times 10^{13}\; \mathrm{N}}$

As both $q_{2}$ and $q_{1}$ are of same charge this force move the charged away and its direction is shown in below figure.

92 = -100 C 91 = -77 C r1 = 2 m F21 41 = -77 C F37 r2 = 3 m Fnet 43 = 50 C

The force exerted by the charge $q_{3}$ on $q_{1}$ can be calculated as follows

$F_{31}=\frac{1}{4\pi \varepsilon _{o}}\frac{q_{1}q_{3}}{r_{2}^{2}}$

Now, substitute the known values

$F_{31}=(9\times 10^{9})\frac{(-77\times 50)}{3^{2}}$

${\color{Blue} F_{31}=-0.385\times 10^{13}\; \mathrm{N}}$

Here the negative sign indicates that the force attractive in nature and its direction is shown in above figure.

The net force on charge $q_{1}$ due to other two charge can be calculated as follows

As both the foces are in same direction we can simply add them

$F_{net}=\left | F_{21} \right |+\left | F_{31} \right |$

$F_{net}=1.7325\times 10^{13}+0.385\times 10^{13}$

${\color{Blue} F_{net}=2.1175\times 10^{13}\; \mathrm{N}}$

Net force direction is shown in above figure

let us replace the charge $q_{3}$ with a test charge

The electric field due to the charge $q_{1}$ on the test charge is

$E_{1}=\frac{1}{4\pi \varepsilon _{o}}\frac{q_{1}}{r^{2}}$

$E_{1}=(9\times 10^{9})\frac{-77}{3^{2}}$

$E_{1}=-77\times 10^{9}\; \mathrm{N/C}$

The direction of this field is shown in below figure

92 = -100 C Enet r1 = 2 m E2 41 = -77 C r2 = 3 m E1 q q

Now, we will calculate the field due to the charge $q_{2}$ on the test charge is

$E_{2}=\frac{1}{4\pi \varepsilon _{o}}\frac{q_{2}}{(r_{2}+r_{1})^{2}}$

$E_{2}=(9\times 10^{9})\frac{-100}{(3+2)^{2}}$

$E_{2}=(9\times 10^{9})\frac{-100}{5^{2}}$

$E_{2}=-36\times 10^{9}\; \mathrm{N/C}$

The direction of this field is shown in below figure

Therefore, the net field on the test charge is

$E_{net}=E_{1}+E_{2}$

$E_{net}=-77\times 10^{9}-36\times 10^{9}$

${\color{Blue} E_{net}=-113\times 10^{9}\; \mathrm{N/C}}$

The direction of the net field is shown in the above figure