Chapter 4: Q27E (page 715)

(a) What must the charge (sign and magnitude) of a 1.45-g particle be for it to remain stationary when placed in a downward-directed electric field of magnitude 650 N/C? (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

Expert verified

Charge of the particle is a negative charge and its magnitude is $q = - 2186 \times 10^{5} C$

Magnitude of the electric field is $1.023 \times 10^{- 7} C$

Step by step solution

Here, (E) is the electric field, (F) is the electric force on charge q, and (q) is the magnitude of the point charge.

Therefore, the electric field is

$E = \frac{F}{q}$

And $F = m g$

Putting the values in

$q = \frac{m g}{E} = \frac{(1.45 \times 10^{- 3}) (9.8)}{650} = 2.186 \times 10^{- 5} C$

As force is upward, so (q) is negative.

Therefore, charge of the particle is a negative charge and its magnitude is $q = - 2186 \times 10^{5} C$ .

As,

$m_{p r o t o n} = 1.6726 \times 10^{- 27} k g q_{p r o t o n} = 1.60217 \times 10^{- 19} C$

So, the equation

$E = \frac{F}{q} F = m g E = \frac{m g}{q}$

Putting values,

$= \frac{(1.6726 \times 10^{- 27}) (9.8)}{1.60217 \times 10^{- 19}} E = 1.023 \times 10^{- 7} C$

Therefore, the Magnitude of the electric field is $1.023 \times 10^{- 7} C$ .

Unlock Step-by-Step Solutions & Ace Your Exams!

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.