Three point charges lie on the x-axis, the first is positive at x=0with a charge

Question

Three point charges lie on the x-axis, the first is positive at x=0with a charge of q1= +5 uC, the second is a negative charge at x=2with a charge q2= -2 uC and the third is negative at x=5 and q3=-10 uC. a) find the magnitude and direction of the force onq2 b) find the magnitude and direction of the electric fieldat the potition of q2.

I used the eq F=(q'q)/(4r^2) and got answers for q1 toq2 and q2 to q3. Do I add them to get the final mag? Isthere a trick to knowing the direction or is it just pointing leftbecause it's being attracted by the left charge and rejected by theright? I'm also having difficulty with the second part of theproblem. The only probs I can find examples on are having todo with net electric fields. Please help.

Explanation / Answer

Part a: Attractive force due to charge q1 at q2 =(9*10^9*2*10^-6*5*10^-6)/2^2 = 2.25*10^-2 N Repulsive force due to charge q3 at q2 = (9*10^9*2*10^-6*10*10^-6)/3^2 = 2*10^-2 N Both forces are acting towards left side. Total magnitude of forces at q2 = 2.25*10^-2 +2*10^-2 =4.25*10^-2 N Part b: Electric Field at position q2 Assume a test charge of +1 C is at position q2. Electric Field due to charge q1 at position q2 istowards the right ( repulsive) = 9*10^9*5*10^-6/2^2=11.25*10^3N/C Electric Field due to charge q3 at position q2 is towards theright ( Attractive) = 9*10^9*10*10^-6/3^2 = 10*10^3 N/C The net Electric is towards the right at positon q2. Total magnitude of Electric Field at position q2 =(11.25+10)*10^3 = 21.24*10^3 N/C

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---