Post

Electric Charges and Forces in an Electric Field

Created by @nathanedwards
 at October 31st 2023, 5:13:51 pm.
AP_Physics_1-Questions
#physics
#electricity
#electric-field

Question:

Two charged particles, particle A and particle B, are placed in an electric field. Particle A experiences an electric force of 5 N, while particle B experiences an electric force of 10 N. If the charge on particle A is +2.5 μC, what is the charge on particle B? (Assume the electric field strength remains constant.)

(A) +1.25 μC

(B) -1.25 μC

(C) +5 μC

(D) -5 μC

Answer:

Let's start by using the equation for electric force:

F=E⋅qF = E \cdot q

where F is the electric force, E is the electric field strength, and q is the charge on the particle.

We are given that particle A experiences an electric force of 5 N, and its charge is +2.5 μC. Simply divide the force by the charge to find the electric field strength for particle A:

EA=FqAE_A = \frac{F}{q_A}
EA=5 N2.5×10−6 CE_A = \frac{5 \, \text{N}}{2.5 \times 10^{-6} \, \text{C}}
EA=2×106 N/CE_A = 2 \times 10^{6} \, \text{N/C}

Now that we know the electric field strength for particle A, we can use it to find the charge on particle B. We are given that particle B experiences an electric force of 10 N. Using the equation for electric force, we can rearrange it to solve for the charge q:

F=EB⋅qBF = E_B \cdot q_B
10 N=(2×106 N/C)⋅qB10 \, \text{N} = (2 \times 10^{6} \, \text{N/C}) \cdot q_B

Solving for q_B, we find:

qB=10 N2×106 N/Cq_B = \frac{10 \, \text{N}}{2 \times 10^{6} \, \text{N/C}}
qB=5×10−6 C=5 μCq_B = 5 \times 10^{-6} \, \text{C} = 5 \, \mu \text{C}

Therefore, the charge on particle B is +5 μC.

The correct answer is (C) +5 μC.

Chat

Save To Bookmark