Two charges are located on the positive x-axis of a coordinate system. Charge q1 = 2.00 ´ 10-9 C, and it is 0.020 m from the origin. Charge q2 = -3.00 ´ 10-9 C, and it is 0.040 m from the origin. What is the electric force exerted by these two charges on a third charge, q3 = 5.00 ´ 10-9 C, located at the origin? (k = 8.99 ´ 109 N·m2/C2)
Answers
Given- q1= 2.00*10^-9C (located at distance of 0.020m from origin) , q2= -3.00*10^-9 C(located at distance of 0.040m from origin) and q3= 5.00*10^-9C (located at on the origin)
To find- The Fe (net electric force )exerted on the third charge q3 due to q1 and q2
Let's see how to tackle this problem
- We have to first find out the Fq3q1( force exerted on q3 due to q1) by using Coulomb's law
- since the nature of the charge q1 is positive so the force will be in opposite direction of q1
Fq3q1= Kq1q3/(r1)^2 (here, r1 = 0.020m)
Now put all the given values
Fq3q1= ( 8.99* 10^9 *2*10^-9*5*10^-9)/(0.020m)^2
we get on doing calculation = 22.4*10^-5N (let say F1)
- Now we have to find out the Fq3q2 ( electric force exerted on q3 due to q2) by using Coulomb's law
- Since the nature of the charge q2 is negative so the force exerted on charge q3 will be towards q2
Fq3q2= Kq2q3/(r2)^2 (here, r2=0.040m)
Put all the given values along with the sign
Fq3q2= ( 8.99*10^9*(-3*10^-9)*5*10^-9)/(0.040m)^2
on calculating we get = -8.4*10^-5 N (let say F2)
- Fnet = √(F1)^2+(F2)^2+2F1F2cosθ
√(22.4*10^-5)^2 + (-8.4*10^-5)^2+2*(22.4*10^-5)*(-8.4*10^-5)Cos180°
√{10^-10 (501.76+70.56+376.32}
10^-5√948.64 = 30.8*10^-5N
The total electric force exerted by the two charges on the third charge q3 is , Fnet = 3.08 *10^-4N Ans