Question

Electricity problem..... please can you solve this...?

(a) What is the magnitude and direction of the electric force exerted on the 5.0 uC charge by the 3.0 uC charge?
(b) What is the magnitude and direction of the electric force exerted on a -4.00 uC charge placed at point X by the other two charges.

Answer 1

`"the solution has shown at explanation section but please check the math operations."`

Explanation:

`a:`

• The electric charge of -3.0 `mu`C pulls the electric charge of 5.0 `mu`C(due to Coulomb's law)) .
• This force has shown as red vector as symbolically.
• The force vector has shown at negative direction.

`"r:6.0 cm=6.10^-2 m (let convert cm to m)"`

`q_1=-3.0 *10^-6 C" (let convert "mu"C to Coulomb)"`

`q_2=+5.0*10^-6 C`

`k=9.10^9 N*m^2*C^-2`

• The magnitude of the force can be calculated using Coulomb's law.

`F=9.10^9(-3.0*10^-6*5.0*10^-6)/((6*10^-2)^2)`

`F=cancel(9).10^9(-3.0*10^-6*5.0*10^-6)/(cancel(36)*10^-4)`

`F=-150/4=-37.5N`

`b:`

• the charge -4 `mu`C pushes by the charge -3 `mu`C toward positive direction(due to Coulomb's law).

• This force has shown as blue vector as symbolically.

`color(blue)(F_1)=9.10^9((-3.10^-6)*(-4*10^-6))/((1.5*10^-2)^2)`

`color(blue)(F_1)=480N`

• the charge -4 `mu`C pulls by the charge -5 `mu`C toward positive direction(due to Coulomb's law).

• This force has shown as green vector as symbolically.

`color(green)(F_2)=9.10^9((-4.10^-6)*(5*10^-6))/((4.5*10^-2)^2)`

`color(green)(F_2)=-8.89N("The negative sign means the force is attractive.")`

• Since vectors `F_1` and`F_2` has the same direction, we must find the vectorial sum of these.

• F=`F_1`+`F_2`

• F=480+8.89=488.89 N