Question

Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x = 40 cm, and –60 µC at x = 60 cm, what is the magnitude of the electrostatic force on the +32-µC charge?

Answer 1

The magnitude of the electrostatic force on
`\mathrm{+32\,\mu C}` is:
`\mathbf{12\;N}`

Step-by-step explanation:

Consider

• 
  `q_1=+32\;\mu\text{C}` is at position
  `x_1=0\;\text{m}`
• 
  `q_2=+20\;\mu\text{C}` is at position
  `x_2=40\;\text{cm}`
• 
  `q_3=-60\;\mu\text{C}` is at position
  `x_3=60\;\text{cm}`

The sum of all horizontal forces on
`q_1` is given as

`\sum F_x=F_(12)+F_(13)`

where

• 
  `F_(12)` is the force exerted by
  `q_2`
• 
  `F_(13)` is the force exerted by
  `q_3`

The force on
`q_1` exerted by
`q_2` is repulsive, so the direction of the force
`F_(12)` is to the left (negative direction). Thus

`F_(12)=-(K\,|q_1|\,|q_2|)/(r_(12)^2)\\F_(12)=-(K\,|q_1|\,|q_2|)/((x_2\;-\;x_1)^2)\\F_(12)=\mathrm+20\;\mu C\\F_(12)=\mathrm{-((8.988*10^9\;(Nm^2)/(C^2))\,(32\;\mu C)\,(20\;\mu C))/((40\;cm)^2)}\\F_(12)=\mathrm{-((8.988*10^9\;(Nm^2)/(C^2))\,(32*10^(-6)\;C)\,(20*10^(-6)\;C))/((0.40\;m)^2)}\\F_(12)=\mathrm{-36\;N}`

The force on
`q_1` exerted by
`q_3` is attractive, so the direction of the force
`F_(13)` is to the right (positive direction). Thus

`F_(13)=+(K\,|q_1|\,|q_3|)/(r_(13)^2)\\F_(13)=+(K\,|q_1|\,|q_3|)/((x_3\;-\;x_1)^2)\\F_(13)=\mathrm-60\;\mu C\\F_(13)=\mathrm{+((9*10^9\;(Nm^2)/(C^2))\,(32\;\mu C)\,(60\;\mu C))/((60\;cm)^2)}\\F_(13)=\mathrm{+((9*10^9\;(Nm^2)/(C^2))\,(32*10^(-6)\;C)\,(60*10^(-6)\;C))/((0.60\;m)^2)}\\F_(13)=\mathrm{+48\;N}`

Therefore

`\sum F_x=\mathrm{-36\;N+48\;N} \;\;\;\;\;\Rightarrow\;\;\;\;\; \mathbf{\sum F_x=+12\;N}`

the electrostatic force on
`q_1` is to the right and has a magnitude of
`\mathbf{12\;N}`