Four charged particles are at the corners of a square of side a, as shown in the figure below. (A = 4, B = 5, C = 5.)
(a) Determine the magnitude and direction of the electric field at the location of charge q. (Use k_e for the Coulomb constant and q and a as necessary.)
Magnitude
______
Direction
______° (counterclockwise from the horizontal)
(b) What is the total electric force exerted on q?
Magnitude
_____
Direction
_____°(counterclockwise from the horizontal)
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I've figured this much:
[(k_e*4q)/(a^2)]i-hat + [(k_e*5q)/(2a^2)](i-hat cos 45 + j-hat sin 45) + (k_e*5q)/(a^2)]j-hat
----> ?
But I don't know how to simplify it from there, though I know my final answer needs to be:
_____*k_e*q)/(a^2)
Oh picture looks like this
4q----a----q
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a a
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5q----a----5q
To simplify the expression, let's break it down step by step.
First, let's look at the x-component of the electric field. The x-component comes from the first term and the second term (the one involving cos 45°).
The first term is [(k_e * 4q) / (a^2)] i-hat, which is the contribution from charges A and B. Since they are on the x-axis, their contribution is entirely in the x-direction.
The second term is [(k_e * 5q) / (2a^2)] (i-hat cos 45°), which is the contribution from charge C. In this term, the cos 45° represents the x-component of the electric field due to the diagonal separation. However, since the diagonal separation is dividing the electric field between charges A and B, its magnitude is divided by 2. Thus, the factor of 2 in the denominator.
To simplify the x-component, add the two terms together:
[(k_e * 4q) / (a^2)] i-hat + [(k_e * 5q) / (2a^2)] (i-hat cos 45°)
Next, let's look at the y-component of the electric field. The y-component comes from the second term (the one involving sin 45°) and the third term.
The second term is [(k_e * 5q) / (2a^2)] (j-hat sin 45°), which is the contribution from charge C. In this term, sin 45° represents the y-component of the electric field due to the diagonal separation. Again, since the diagonal separation is dividing the electric field between charges A and B, its magnitude is divided by 2.
The third term is [(k_e * 5q) / (a^2)] j-hat, which is the contribution from charge D. Its contribution is entirely in the y-direction.
To simplify the y-component, add the two terms together:
[(k_e * 5q) / (2a^2)] (j-hat sin 45°) + [(k_e * 5q) / (a^2)] j-hat
Now, the total electric field is the sum of the x-component and the y-component:
E = [(k_e * 4q) / (a^2)] i-hat + [(k_e * 5q) / (2a^2)] (i-hat cos 45°) + [(k_e * 5q) / (2a^2)] (j-hat sin 45°) + [(k_e * 5q) / (a^2)] j-hat
To simplify this further, let's factor out the common terms:
E = [(k_e * 4q) / (a^2)] i-hat + [(k_e * 5q) / (a^2)] (i-hat cos 45° + j-hat sin 45°) + [(k_e * 5q) / (2a^2)] j-hat
Now, notice that i-hat cos 45° + j-hat sin 45° can be rewritten as the unit vector pointing at a 45° angle. This unit vector is given by √2/2 (i-hat + j-hat), since cos 45° = sin 45° = √2/2.
Substituting this into the expression for the electric field:
E = [(k_e * 4q) / (a^2)] i-hat + [(k_e * 5q) / (a^2)] (√2/2) (i-hat + j-hat) + [(k_e * 5q) / (2a^2)] j-hat
Now, let's simplify further. Distribute the terms:
E = [(k_e * 4q) / (a^2)] i-hat + [(k_e * 5q * √2) / (2a^2)] i-hat + [(k_e * 5q * √2) / (2a^2)] j-hat + [(k_e * 5q) / (2a^2)] j-hat
Combine like terms:
E = [(k_e * (4q + 5q * √2)) / (a^2)] i-hat + [(k_e * (5q * √2 + 5q)) / (2a^2)] j-hat
Simplify the coefficients:
E = [(k_e * (4q + 5q * √2)) / (a^2)] i-hat + [(k_e * (5q * (1 + √2))) / (2a^2)] j-hat
Now, compare this expression to the desired final answer: [(k_e * q)/(a^2)].
From the comparison, we can deduce the magnitude of the electric field at the location of charge q:
Magnitude = √[(4q + 5q * √2)^2 + (5q * (1 + √2))^2] / (a^2)
To determine the direction, we need to find the angle counterclockwise from the horizontal.
Direction = arctan[(5q * (1 + √2)) / (4q + 5q * √2)]
For part (b), the total electric force exerted on charge q is given by:
F = q * E
where E is the electric field we just calculated. Substitute the expression for E and multiply by q to find the magnitude of the force. To find the direction, use the same angle you calculated for the electric field.
Magnitude = q * √[(4q + 5q * √2)^2 + (5q * (1 + √2))^2] / (a^2)
Direction = arctan[(5q * (1 + √2)) / (4q + 5q * √2)]