Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105

Question

5 days ago

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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.

Physics

2 answers:

Keith_Richards [2.2K] · Answer 1 · 2026-03-20T23:25:22+03:00

The particle’s charge q₁ is approximately 7.25 × 10⁻¹⁶ C

More details

Electric charge can be classified into two categories: positive and negative charges.

The renowned scientist Coulomb conducted significant research on electric charges, resulting in the formulation of the forces of attraction and repulsion between them. He expressed this with the equation:

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

F = electric force (N)

k = electric constant (N m² / C²)

q = electric charge (C)

r = distance between charges (m)

In a vacuum, k is valued at 9 x 10⁹ (N m² / C²)

Now, let’s address the problem!

Provided:

diameter of the 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 =?

Resolution:

The drag force arises from the electric force acting on the charged 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:
A series circuit:
Contrast and compare series and parallel circuits:

Answer specifics

Grade: High School

Subject: Physics

Chapter: Static Electricity

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

Ostrovityanka [2.2K] · Answer 2 · 2026-03-20T23:25:42+03:00

This is somewhat misleading, and I encountered the same question in my homework. An electric field strength of 1*10^5 N/C is provided, along with a drag force of 7.25*10^-11 N, and the critical detail is that it maintains a constant velocity, indicating that the particle is in equilibrium and not accelerating.
To solve, utilize F=(K*Q1*Q2)/r^2 
You'll want to equate F with the drag force, where the electric field strength translates to (K*Q2)/r^2; substituting the values results in 
(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C