Question

Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105 newtons per coulomb while acted upon by a drag force of 7.25×10−11 newtons. what is the charge q1 on the particle? ignore the effects of gravity.

Answer 1

The charge q₁ on the particle is about 7.25 × 10⁻¹⁶ C

Further explanation

Electric charge consists of two types i.e. positively electric charge and negatively electric charge.

There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :

$\large {\boxed {F = k \frac{Q_1Q_2}{R^2} } }$

F = electric force (N)

k = electric constant (N m² / C²)

q = electric charge (C)

r = distance between charges (m)

The value of k in a vacuum = 9 x 10⁹ (N m² / C²)

Let's tackle the problem now !

Given:

diameter of charged particle = d = 1 μm

electric field strength = E = 1.00 × 10⁵ N/C

drag force = F = 7.25 × 10⁻¹¹ N

Unknown:

charge of particle = q₁ = ?

Solution:

The drag forces is caused by the electric force acting on the charge particles.

$F = q_1 \times E$

$7.25 \times 10^{-11} = q_1 \times 1.00 \times 10^5$

$q_1 = (7.25 \times 10^{-11}) \div (1.00 \times 10^5)$

$q_1 = 7.25 \times 10^{-16} ~ \text{Coulomb}$

Learn more

• The three resistors : brainly.com/question/9503202
• A series circuit : brainly.com/question/1518810
• Compare and contrast a series and parallel circuit : brainly.com/question/539204

Answer details

Grade: High School

Subject: Physics

Chapter: Static Electricity

Keywords: Series , Parallel , Measurement , Absolute , Error , Combination , Resistor , Resistance , Ohm , Charge , Small , Forces

Answer 2

It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating). 
All you need is F=(K*Q1*Q2)/r^2 
Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you 
(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C