Question

You want to create an electric field vector E = < 0, 5 104, 0> N/C at location < 0, 0, 0>. Where would you place a proton to produce this field at the origin?

Answer 1

Answer:

1.696 × 10^(-7) m on the y axis.

Explanation:

We are given the electric field as;

E = < 0, 5 × 10⁴, 0> N/C

This is written in (x, y, z) co-ordinates. So it means that it lies on the y-axis.

So,

E = 5 × 10⁴ N/C in the y direction.

Formula for Electric field is;

E = kq/r²

where;

k is a constant with a value of 8.99 x 10^(9) N.m²/C²

q is charge on the proton = 1.6 × 10^(-19) C

r is the distance

Thus, making r the subject gives;

r = √(kq/E)

Plugging in the relevant values gives;

r = √(8.99 × 10^(9) × 1.6 × 10^(-19)/(5 × 10⁴))

r = 1.696 × 10^(-7) m on the y axis.

Answer 2

The proton to produce this field at the origin should be put on $1.696 \times 10^{(-7)}$m on the y axis.

• The calculation is as follows:

We are given the electric field as;

$E = < 0, 5 \times 10^{4}, 0> N/C$

This is written in (x, y, z) co-ordinates. So it means that it lies on the y-axis.

So,

$E = 5 \times 10^{4} N/C$ in the y direction.

The Formula for Electric field is;

E = kq/r²

where;

k is a constant with a value of 8.99 x 10^(9) N.m²/C²

q is charge on the proton = 1.6 × 10^(-19) C

r is the distance

Thus, making r the subject gives;

r = √(kq/E)

Now Plugging in the relevant values gives;

r = √(8.99 × 10^(9) × 1.6 × 10^(-19)/(5 × 10⁴))

r = 1.696 × 10^(-7) m on the y axis.

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