A car moves along a straight path with a speed of 30kmh. How much distance will the car travel in 20 minutes
1 answer:
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Answer:
Er = 231.76 V/m, 27.23° to the left of E1
Explanation:
To find the resultant electric field, you can use the component method. Where you add the respective x-component and y-component of each vector:
E1:
E2:
Keep in mind that the x component of electric field E2 is directed to the left.
∑x:
∑y:
The magnitud of the resulting electric field can be found using pythagorean theorem. For the direction, we will use trigonometry.
or 27.23° to the left of E1.
Answer:
For situation (a)
net charge E = E₊₂ + E₋₅ + E₋₃
E = K(q/d²)
where K = 8.99e9
d = 5.7cm = 5.7e-2m
Therefore,
E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)
E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)
E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)
thus
E = E₊₂ + E₋₅ + E₋₃
= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)
= 7.88e6(x) + 7.88e6(y)
use Pythagorean theorem
I <em>E </em>I = = 1.242e6
∅ = =
= 45°
Thus for (a) net magnitude = 1.115e6 @ 45° above +x axis
for situation (b)
net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆
E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)
E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)
E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)
E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)
thus,
E = E₊₄ + E₊₁ + E₋₁ + E₊₆
= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)
= 7.88e5(x) + 7.88e5(y)
use Pythagorean theorem
I <em>E </em>I = = 1.242e6
∅ = =
= 45°
Thus for (a) and (b) the net magnitude = 1.242e6 @ 45° above +x axis
Explanation:
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