The answer is B. Please explain to me how to asses positive and negative signs a

Question

The answer is B. Please explain to me how to asses positive and negative signs and when do you assume charges in absolute value? Thank you so much

Three changes are located along the x axis as shown in the drawing. The mass of the -1.2 muC is 1.5 Times 10-9 kg. Determine the magnitude and direction of the acceleration of the -1.2 muC charge when it is allowed to move if the other two changes remain fixed. 2.0 Times 105 m/s2 , to the left 1.0 Times 105 m/s2 , to the right 7.0 Times 105 m/s2 , to the right 3.0 Times 105 m/s2 , to the left 4.0 Times 105 m/s2 , to the right

Explanation / Answer

here

q1 = 1.5 * 10^-6 C

q2 = -1.2 * 10^-6 C

q3 = 2 * 10^-6 C

first we find the force on q2 due to q1 is

F1 = k * q1 * q2 / r^2

F1 = 9 * 10^9 * 1.5*10^-6 * 1.2 * 10^-6 / 6^2

F1 = 0.45 * 10^-3 N

the force due to q1 charge on q2 charge is 0.45 * 10^-3 N in the direction toward the q1 charge in the -ve xaxis and there is force of attraction

then we find the force due to q3 on q2

F2 = 9 * 10^9 * 1.2 * 10^-6 * 2 * 10^-6 / 6^2

F2 = 0.6 * 10^-3 N

the force due to q3 charge on q2 charge is 0.6 * 10^-3 N in the direction toward the q1 charge in the +ve xaxis and there is force of attraction

so the total force on the q2 is

F = F2 - F1

F = (0.6 - 0.45 ) * 10^-3

F = 0.15 * 10^-3 N

so the force on the charge q2 is 0.15 * 10^-3 in the direction of +ve axis or toward the q3 charge

then

by using

F = m* a

a = 0.15 * 10^-3 / 1.5 * 10^-9

a = 0.1 * 10^-6 = 1 * 10^-5 m/s^2

so the acceleration of the q2 charge is 1 * 10^-5 m/s^2 to the right

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