Question

Consider two point charges located on the x axis: one charge, q1 = -11.5nC , is located at x1 = -1.685m ;the second charge, q2 = 40.0nC , is at the origin (x=0.0000).What is the net force exerted by these two charges on a third charge q3 = 55.0nC placed between q1 and q2 at x3 = -1.195m ?

Answer 1

To find the net force exerted by the two charges on the third charge, we can use Coulomb's Law. The net force exerted by the two charges on the third charge is approximately 8.17 x 10-8 N, directed towards charge q1.

Step-by-step explanation:

To find the net force exerted by the two charges on the third charge, we can use Coulomb's Law:

F = k * q1 * q3 / r132 + k * q2 * q3 / r232

Where:

• F is the net force between the charges
• k is the electrostatic constant (8.99 x 109 Nm2/C2)
• q1 and q2 are the charges of q1 and q2 respectively
• q3 is the charge of q3
• r13 and r23 are the distances between q1 and q3 and between q2 and q3 respectively

Plugging in the values:

• F = (8.99 x 109 Nm2/C2) * (-11.5nC) * (55.0nC) / ((-1.685m - (-1.195m))2) + (8.99 x 109 Nm2/C2) * (40.0nC) * (55.0nC) / ((0.0000m - (-1.195m))2)

The net force exerted by the two charges on the third charge is approximately 8.17 x 10-8 N, directed towards charge q1.

Answer 2

`F_(net) = 3.75*10^(-5) Newton towards negative axis`

Step-by-step explanation:

q1 = -11.5nC , x1 = -1.685m

q2 = 40.0nC , x=0.0000m

q3 = 55.0nC , x3 = -1.195m

Force on charge 3 due to charge 1 (attractive)

By applying coulomb law

`F_(13)=K(q_(1)q_(3))/(r^(2) )`

r = 1.685-1.195

r = 0.49 m

`F_(13)=9*10^(9)* (-11.5nC * -55.0nC)/(0.49^(2)m ) * -\iota`

`F_(13)= 2.37*10^(-5) (-\iota) N`

Force on charge 3 due to charge 2 (repulsive)

`F_(23)=K(q_(2)q_(3))/(r^(2) )`

`F_(23)=9*10^(9)* (-40nC * -55.0nC)/(1.195 ^(2)m )`

`F_(23)=1.38* 10^(-5)* -\iota N`

`F_(net) =F_(13) +F_(23)`

`F_(net) = 3.75*10^(-5) Newton   towards negative axis`