Question

Consider three charges arranged as shown. Sphere A has a charge of 6.40 x 10-6 coulombs. Sphere B has a charge of 6.51 x 10-6 coulombs and is 0.417 meters from sphere A. Sphere C has a charge of 2.12 x 10-6 coulombs and is 0.301 meters from sphere A.a. What is the magnitude of the force sphere B exerts on sphere A? Include units in your answer.

Answer 1

Given:

The charge on sphere A, q_A=6.40×10⁻⁶ C

The charge on sphere B, q_B=6.51×10⁻⁶ C

The charge on sphere C, q_C=2.12×10⁻⁶ C

The distance between charge A and B, d_B=0.417 m

The distance between charge A and C, d_C=0.301 m

To find:

a. Magnitude of force between A and B.

b. The magnitude of the force between A and C.

c. The net force on A.

d. The magnitude of the electric field at A.

Step-by-step explanation:

The force between two charges is given by,

`F=(kq_1q_2)/(d^2)`

Where k is the coulomb's constant, q₁ and q₂ are the charges and d is the distance between the two charges.

a.

The force between A and B is,

`\begin{gathered} F_(AB)=(kq_Aq_B)/(d_B^2) \\ =(9*10^9*6.40*10^(-6)*6.51*10^(-6))/(0.417^2) \\ =2.16\text{ N} \end{gathered}`

b.

The force between A and C is,

`\begin{gathered} F_(AC)=(kq_Aq_C)/((d_C)^2) \\ =(9*10^9*6.40*10^(-6)*2.12*10^(-6))/(0.301^2) \\ =1.35\text{ N} \end{gathered}`

c.

The forces when represented in the vector form,

`\begin{gathered} \vec{F_(AB)}=2.16(-\hat{i}) \\ \vec{F_(AC)}=1.35(-\hat{j}) \end{gathered}`

Where i_cap and j_cap are the unit vectors along the negative x-axis and negative y-axis respectively.

The net force acting on sphere A is,

`\vec{F_A}=\vec{F_(AB)}+\vec{F_(AC)}`

On substituting the known values,

`\vec{F_A}=-2.16\hat{i}-1.35\hat{j}`

The magnitude of the net force is given by,

`\begin{gathered} F_A=√((-2.16)^2+(1.35)^2) \\ =2.55\text{ N} \end{gathered}`

d.

The electric field at point A due to charge B is

`\vec{E}_B=(kq_B)/(d_B^2)(-\hat{i})`

On substituting the known values,

`\begin{gathered} \vec{E_B}=(9*10^9*6.51*10^(-6))/(0.417^2)(-\hat{i}) \\ =-336.9*10^3\hat{i}\text{ N/C} \end{gathered}`

The electric field at point A due to charge C is

`\vec{E}_C=(kq_C)/((d_C)^2)(-\hat{j})`

On substituting the known values,

`\begin{gathered} \vec{E_C}=(9\cdot10^9*2.12*10^(-6))/(0.301^2)(-\hat{j}) \\ =-210.6*10^3\hat{j}\text{ N/C} \end{gathered}`

The electric field at point A due to sphere A is zero ans the distance from sphere A to point A is zero

Thus the net electric field at point A is given by,

`\vec{E}=-336.9*10^3\hat{i}-210.6*10^3\hat{j}`

The magnitude of the net electric at point A is,

`\begin{gathered} E=√((-336.9*10^3)^2+(-210.6*10^3)^2) \\ =397.3*10^3\text{ N/C} \end{gathered}`

Step-by-step explanation:

a. The magnitude of the force exerted on sphere A by sphere B is 2.16 N

b. The magnitude of the force exerted on sphere A by sphere C is 1.35 N

c. The magnitude of the net force on sphere A is 2.55 N

c. The magnitude of the net electric field at point A is 397.3×10³ N/C