Question

A particle (charge = +40 μC) is located on the x axis at the point x = –20 cm, and a second particle (charge = –50 μC) is placed on the x axis at x = +30 cm. What is the magnitude of the total electrostatic force on a third particle (charge = –4.0 μC) placed at the origin (x = 0)?

Answer 1

Answer:

56N

Explanation:

The magnitude of the total force on the third particle is equal to the resultant force on it due to the first and second particles. From Coulomb's law of electrostatic attraction, the following relationship holds;

$F=\frac{kq_1q_2}{r^2}...................(1)$

where k is the electrostatic constant whose value is 9 x $10^9Nm^2/C^2$ and

r is the distance between the two charges $q_1$ and $q_2$.

According to the problem given, let the first charge be $q_1$, the second charge be $q_2$ and the third charge at the origin be $q_o$. Therefore;

$q_1=+40\mu C\\q_2=-50\mu C\\q_o=-4\mu C$

Also, let the force due to the first charge be $F_1$ and the force due to the second charge be $F_2$.

Hence, by equation (1); we can write the following;

$F_1=\frac{kq_1q_o}{r_1^2}............(2)$

$F_2=\frac{kq_2q_o}{r_1^2}............(3)$

where $r_1=20cm =0.2m$ and $r_2=30cm=0.3m$

$F_1$ is attractive since $q_1$ and $q_o$ are unlike charges while $F_2$ is repulsive since $q_2$ and $q_o$ are like charges.

Substituting all values into equations (2) and (3) we obtain the following;

$F_1=\frac{9*10^9*40*10^{-6}*4*10^{-6}}{0.2^2}=36N$

$F_2=\frac{9*10^9*50*10^{-6}*4*10^{-6}}{0.3^2}=20N$

Both $F_1$ and $F_2$ point towards the left (see attached diagram), hence the resultant force on $q_o$ becomes

$F=F_1+F_2$

F = 36 +20

F = 56N