Question

In: Physics

A doubly‑ionized carbon atom (with charge +2e) is located at the origin of the x‑axis, and...

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=8.76 cm.There is one location along the x‑axis at which the electric field is zero. Give the x‑coordinate of this point in centimeters.

x‑coordinate:

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 the two points along the x‑axis at which the potential is zero, and express their locations along the x‑axis in centimeters, starting with the point that is farther away from the origin.

x‑coordinate of the farther point:

cm

x‑coordinate of the closer point:

cm

Expert Solution

The formula for the electric field is E = kq/r²

For the field to be zero, the individual fields from the two charges must cancel. That location can be found by

kq/r² = kq/r² (Note k cancels and the value of q, we can use 1 and 2 to represent the charges)

Call the distance to the point x from the negative charge, and it will be to the right since the negative charge is lower in magnitude than the positive charge. Then the distance from the positive charge is x + .0876

1/x² = 2/(x +.0876)²

Take the square root of everything and you get

1/x = √2/(x + .0876)

Cross multiply

x + .0876 = 1.414x

.0876 = .414x

x = .2115 m

Since that distance is from the negative charge, the location on the x axis is at .299 m which is 29.9 cm

Part B)

Potential is a scalar quantity and adds algebraically. The formula for potential is

V = kq/r

Thus kq/r = kq/r for the potentials to cancel (again k cancels and the charges can be represented as 1 and 2)

1/x = 2/x + .0876 (use x as the distance from the negative charge)

x = .0876

The distance is therefore .0876+ .0876. Thus at 17.52 cm will be one location for zero potential

1/x = 2/.0876 - x

x = .0142

The other distance is .0876 - .0142. Thus at 7.34 cm we will also have zero potential

So, the closer point is 7.34 cm and the farther point is 17.52 cm

genius_generous answered 2 years ago

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