A 7.00 microcoulomb charge is .500 m above a 15.0 microcoulomb charge. Calculate the electric field at a point 1.00m above the 15.0 microcoulomb charge.
Charges of -3.50 nC and -7.00 nC are 20.0cm apart. Find a 5.0 nC charge’s equilibrium position.
- billrussell42Lv 71 month ago
E fields add as vectors
assuming both charges are +
from the first charge
E = k7µ/0.5² = 28e-6(9e9) V/m pointed up
E = 252 kV/m pointed up
from the second chargeE = k15µ/1² = 15e-6(9e9) V/m pointed upE = 135 kV/m pointed up
sum = 387 kV/m pointed up
The strength or magnitude of the field at a given point
is defined as the force that would be exerted on apositive test charge of 1 coulomb placed at that point;
the direction of the field is given by the direction ofthat force.
E = F/Q = kQ/r²
Q = F/E F = QE
in Newtons/coulomb OR volts/meter
k = 1/4πε₀ = 8.99e9 Nm²/C²