Electric potential and electric field

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SUMMARY

The discussion focuses on calculating the electric field at a point on the x-axis between a -10.0 nC and a +20.0 nC point charge, which are 15 cm apart. The electric potential is zero at a distance of 0.1 m from the -10.0 nC charge. The calculated electric field at this point is 18,000 V/m. The user expresses uncertainty about the direction of the electric fields from both charges and seeks clarification on the correct approach to summing them.

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[SOLVED] Electric potential and electric field

Homework Statement



A - 10.0 nC point charge and a + 20.0nC point charge are 15 cm apart on the x-axis.
What is the magnitude of the electric field at the point on the x-axis, between the charges, where the electric potential is zero?

Homework Equations


sumE = E(of -10) + E (of 20)
E = kq/r^2
SumV = V(-10) +V(20)
V = kq/r


The Attempt at a Solution



let the distance at which V = 0 between the charge be x

SumV = V(-10) +V(20) = 0
k { (-10 *10^-9/0.15 - x) + (20 *10^-9/ x)} = 0
solve for x
x = 0.1
0.15 - x = 0.05

SumE = E(-10) +E(20)
k { (-10 *10^-9/(0.15 - x)^2) + (20 *10^-9/ (x)^2)} = SumE
substitute x = 0.1
Sum E= 18000 V

i don't think this is the right answer
i think i am doing something wrong
 
Physics news on Phys.org
The directions of the electric field due to two charges is the same. So the net field is the sum of the absolute values of the two fields.
 
thank you
 

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