Two point charges lie on the x axis. a charge of + 2.20 pc is at the origin, and a charge of − 4.80 pc is at x=−11.0cm. what third charge should be placed at x=+20cm so that the total electric field at x=+10.0cm is zero?

The charge must be -1.11pC

Explanation: The field done by a charge is equal to:

E(x) = k*q/(x - x')^2

Where k is the constant, q is the charge and x' is the position of the charge, then, the total field at the point x = 10cm must be also zero, we have:

E(10cm) = k*(2.20pC/(10cm - 0cm)^2 - 4.80pC/(10cm + 11cm)^2 + Q/(10cm - 20cm)^2) = 0

And we must find the value of Q.

(2.20pC/(10cm)^2 - 4.80pC/(21cm)^2 + Q/(-10cm)^2) = 0

Q = (-2.20pC/(10cm)^2 + 4.80pC/(21cm)^2)*(10cm)^2 = -1.11pC

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm