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Math  /  Algebra

QuestionA point charge of 2μC2 \mu \mathrm{C} is placed at the origin. There is an external uniform field E=600iN/C\mathbf{E}=600 \mathrm{iN} / \mathrm{C}. What is the net force on a 6μC6 \mu \mathrm{C} charge placed at (3 m,3 m)(3 \mathrm{~m}, 3 \mathrm{~m}) ? 7.843⋅10−3j+4.243⋅10−3jN7.843 \cdot 10^{-3} \mathrm{j}+4.243 \cdot 10^{-3} \mathrm{jN}
Note: You can earn partial credit on this problem. Show: □\square Correct Answers

Studdy Solution

STEP 1

What is this asking? Find the total force on a charge due to both a point charge at the origin and a uniform electric field. Watch out! Don't forget to consider *both* the force due to the point charge *and* the force due to the external field!
Also, remember those vector components!

STEP 2

1. Calculate the force due to the point charge.
2. Calculate the force due to the electric field.
3. Find the total force.

STEP 3

We'll use Coulomb's Law to find the force between the two charges.
Coulomb's Law says the force between two charges is: F=k⋅q1⋅q2r2F = k \cdot \frac{q_1 \cdot q_2}{r^2} where kk is Coulomb's constant (k≈8.988⋅109 N⋅m2/C2k \approx 8.988 \cdot 10^9 \ \text{N} \cdot \text{m}^2/\text{C}^2), q1q_1 and q2q_2 are the magnitudes of the charges, and rr is the distance between them.

STEP 4

The first charge is at the origin (0,0)(0, 0) and the second charge is at (3 m,3 m)(3 \ \text{m}, 3 \ \text{m}).
The distance rr between them is: r=(3−0)2+(3−0)2=32+32=18=32 mr = \sqrt{(3 - 0)^2 + (3 - 0)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \ \text{m}

STEP 5

Now, plug everything into Coulomb's Law: F=(8.988⋅109)⋅(2⋅10−6)⋅(6⋅10−6)(32)2=(8.988⋅109)⋅12⋅10−1218≈5.992⋅10−3 NF = (8.988 \cdot 10^9) \cdot \frac{(2 \cdot 10^{-6}) \cdot (6 \cdot 10^{-6})}{(3\sqrt{2})^2} = (8.988 \cdot 10^9) \cdot \frac{12 \cdot 10^{-12}}{18} \approx 5.992 \cdot 10^{-3} \ \text{N}

STEP 6

Since both charges are positive, the force is repulsive, pointing away from the origin.
The direction of the force is the same as the direction of the vector pointing from the origin to (3,3)(3, 3), which makes a **45-degree angle** with both the x and y axes.

STEP 7

The x and y components of the force are equal since the angle is 45 degrees.
We can find the components using: Fx=F⋅cos⁡(45∘)=5.992⋅10−3⋅22≈4.237⋅10−3 NF_x = F \cdot \cos(45^\circ) = 5.992 \cdot 10^{-3} \cdot \frac{\sqrt{2}}{2} \approx 4.237 \cdot 10^{-3} \ \text{N} Fy=F⋅sin⁡(45∘)=5.992⋅10−3⋅22≈4.237⋅10−3 NF_y = F \cdot \sin(45^\circ) = 5.992 \cdot 10^{-3} \cdot \frac{\sqrt{2}}{2} \approx 4.237 \cdot 10^{-3} \ \text{N}

STEP 8

The force on a charge qq in an electric field E\mathbf{E} is given by: F=qE\mathbf{F} = q\mathbf{E} Here, q=6⋅10−6 Cq = 6 \cdot 10^{-6} \ \text{C} and E=600i N/C\mathbf{E} = 600\mathbf{i} \ \text{N/C}.

STEP 9

F=(6⋅10−6)⋅(600i)=3.6⋅10−3i N\mathbf{F} = (6 \cdot 10^{-6}) \cdot (600\mathbf{i}) = 3.6 \cdot 10^{-3}\mathbf{i} \ \text{N} So, Fx=3.6⋅10−3 NF_x = 3.6 \cdot 10^{-3} \ \text{N} and Fy=0F_y = 0.

STEP 10

Now, we add the x and y components of the forces we calculated: Fx,total=4.237⋅10−3+3.6⋅10−3=7.837⋅10−3 NF_{x, \text{total}} = 4.237 \cdot 10^{-3} + 3.6 \cdot 10^{-3} = 7.837 \cdot 10^{-3} \ \text{N} Fy,total=4.237⋅10−3+0=4.237⋅10−3 NF_{y, \text{total}} = 4.237 \cdot 10^{-3} + 0 = 4.237 \cdot 10^{-3} \ \text{N}

STEP 11

The total force is: Ftotal=7.837⋅10−3i+4.237⋅10−3j N\mathbf{F}_{\text{total}} = 7.837 \cdot 10^{-3}\mathbf{i} + 4.237 \cdot 10^{-3}\mathbf{j} \ \text{N}

STEP 12

The net force on the 6 μC6 \ \mu\text{C} charge is 7.837⋅10−3i+4.237⋅10−3j N7.837 \cdot 10^{-3}\mathbf{i} + 4.237 \cdot 10^{-3}\mathbf{j} \ \text{N}.

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