A. What is the magnitude of the resulting force on the -7 nC charge at the origi

Question

A. What is the magnitude of the resulting force on the -7 nC charge at the origin?

B. What angle does the resultant force on -7  * 10^-9 C charge at the origin make with the positive x axis?  

Here is the plane:  http://imgur.com/7aM0l9l

I've tried finding the magnitudes of 4nC and 8nC and then mulplying them by their respective sin(?) and cos(?).  Then I was using the pythagorean theorum to find the Fnet but I cannot seem to figure out if that is the correct way.

What is the magnitude of the resulting force on the -7 nC charge at the origin? What angle does the resultant force on -7 * 10^-9 C charge at the origin make with the positive x axis?

Explanation / Answer

a)

Fnet = F1 +F2

F1 = kqQ/r^2 = 9*10^9*8*10^-9*7*10^-9/(2^2+5^2) = 1.74*10^-8 N <---this is the magnitude

direction = atan(5/2) in degrees = 68.2 degrees

So, F1 = 1.74*10^-8*(cos(68.2) i + sin(68.2) j)

F2 = 9*10^9*4*10^-9*7*10^-9/(4^2+1^2) = 1.48*10^-8 <---magnitude

direction = atan(1/4) in degrees = 14.04 degrees

so, F2 = 1.48*10^-8*(cos(14.04) i + sin(14.04) j)

So, Fnet = F1+F2= 1.74*10^-8*(cos(68.2) i + sin(68.2) j) + 1.48*10^-8*(cos(14.04) i + sin(14.04) j)

So, Fnet = (1.74*10^-8*cos(68.2) +1.48*10^-8*(cos(14.04)) i + (1.74*10^-8*sin(68.2) +1.48*10^-8*sin(14.04)) j

= 2.1*10^-8 i + 1.97*10^-8

So, magnitude of Fnet = sqrt(2.1^2+1.97^2)*10^-8

= 2.88*10^-8 N <-----------answer

b)

angle = atan(1.97/2.1) in degrees = 43.2 degrees <--------answer

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