A doubly ionized carbon atom (with charge 2e) is located at the origin of the x

Question

A doubly ionized carbon atom (with charge 2e) is located at the origin of the x axis, and an electron (with charge –e) is placed at x = 2.01 cm. a. There is one location along the x axis at which the electric

There is one location along the x axis at which the electric field is zero. Give the x coordinate of this point in cm.

Assume that the potential is defined to be zero infinitely far away from the particles. Unlike the electric field, the potential will be zero at multiple points near the particles. Find two Points along the x axis at which the potential is zero, and express their locations along the x axis in cm, starting with the point which is farther away from the origin.

Explanation / Answer

1) Let q1 = +2*e

q2 = -e

here, |q2| < |q1|, so the point where net electric filed is close to q2 and it is right of q2.

let x is the distance from q2 where Enet = 0

Apply, E1 = E2

k*q1/(2.01 + x)^2 = k*q2/x^2

2*e/(2.01 + x)^2 = e/x^2

x^2/(2.01+x)^2 = 1/2

x/(2.01+x) = sqrt(1/2)

x = 0.707*(2.01+x)

x*(1 - 0.707) = 0.707*2.01

x = 0.707*2.01/(1 - 0.707)

= 4.85 cm

so, on x axis at x = 2.01 + 4.85 = 6.86 cm Enet = 0 <<<<<-----Answer

But the point where Vnet is zero is close to q2 but it is left of q2.

let x is the distance from q2 where Vnet = 0

V1 = V2

k*q1/(2.01-x) = k*q2/x

2*e/(2.01-x) = e/x

2*x = 2.01 - x

3*x = 2.01

x = 2.01/3

= 0.67 cm

so, at x = 2.01 - 0.67 = 1.34 cm Vnet = 0 <<<<<-----Answer

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