Question

Three tiny charged metal balls are arranged on a straight line. The middle ball is positively charged and the two outside balls are negatively charged. The two outside balls are separated by 20 cm and the middle ball is exactly halfway in between. (HINT: Draw a picture; in your picture, the distance between the two outermost balls should be 20 cm.) The absolute value of the charge on each ball is the same, 1.45 μCoulombs (the meaning of μ, which is read as "micro", is 10-6). Give your answers in newtons.

a) What is the magnitude of the attractive force on either outside ball due ONLY to the positively-charged middle ball?

(b) What is the magnitude of the repulsive force on either outside ball due ONLY to the other outside ball?

(c) What is the magnitude of the net force on the outside of the ball

Answer 1

`(a) 189.23 N`,
`(b) 47.31 N` and
`(c) 141.92 N`.

Step-by-step explanation:

Three balls are shown in figure having charge
`q=1.45 \mu C`. The middle ball,
`B`, is positively charged having charge
`+q`, and the remaining two outside balls,
`A` and
`C`, are negatively charged having charged
`-q` as shown.

`AC=20 cm` and
`AB=BC=10` cm as B is the mid-point of AC.

Let
`d_1=AC=20* 10^(-3)m` and
`d_2=AB=BC=10* 10^(-3)m`

From Coulomb's law, the magnitude of the force,
`F`, between two point charges having magnitudes
`q_1 \& q_2`, separated by distance,
`d`, is

`F=\frac {1}{4\pi\epsilon_0}\frac {q_1q_2}{d^2}\;\cdots (i)`

where,
`\epsilon_0` is the permittivity of free space and

`\frac {1}{4\pi\epsilon_0}=9* 10^9` in SI units.

This force is repulsive for the same nature of charges and attractive for the different nature of charges.

Now, Using equations(i),

(a) The magnitude of attraction force between balls A and B is

`F_(AB)=F_(BC)= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_2)^2}`

`\Rightarrow F_(AB)= 9* 10^9}\frac {1.45* 10^(-6)*1.45* 10^(-6)}{\left(10* 10^(-3)\right)^2}`

`\Rightarrow F_(AB)=189.23 N`

(a) The magnitude of the repulsive force between balls A and C is

`F_(AC)= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_1)^2}`

`\Rightarrow F_(AC)= 9* 10^9}\frac {1.45* 10^(-6)*1.45* 10^(-6)}{\left(20* 10^(-3)\right)^2}`

`\Rightarrow F_(AC)=47.31 N`

(c) The magnitude of the net force,
`F_(net)`, on the outside of the ball is,

`F_(net)=189.23-47.31 N`

`\Rightarrow F_(net)=141.92 N`