What is the x-component of the electric field at (x, y) = (0 m, 4 m)??

1 answer:

8 0

(a)Two charges are placed on the x-axis: one is placed at x = 3 m and the other is at x = -3 m. The magnitude of both charges is 2.3 µC but the blue one (at x = -3 m) is positive while the red one (at x = +3 m) is negative.

What are the x- and y-components of the electric field at (x, y) = (0 m, +4 m)?

(b) Now the positive and negative charge switch places. The magnitude of the charges is still 2.3 µC where the blue one (now at x = +3 m) is positive and the red one (now at x = -3 m) is negative.

What are the x- and y-components of the electric field at (x, y) = (0 m, +4 m)?

(c) Now both charges (still at x = -3 m and x = +3 m) are positive. The magnitude of both charges is still 2.3 µC.

What are the x- and y-components of the electric field at (x, y) = (0 m, +4 m)?

(d) Finally, both charges (still at x = -3 m and x = +3 m) are negative. The magnitude of both charges is still 2.3 µC.

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NikAS [45]

Hey there! :)

Answer:

a) 24 cm²

b) 40.04 cm²

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Use the formula A = l × w to solve for each rectangle's area:

8 × 3 = 24 cm².

5.2 × 7.7 = 40.04 cm²

6 0

3 years ago

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Suppose a railroad rail is 3 kilometers and it expands on a hot day by 12 centimeters in length. Approximately how many meters w

inysia [295]

For small angles, x, in radians,
tan(x) ≈ sin(x) = x

Use this to create an approximate diagram for the problem as shown in the figure below.
The actual deflected shape is parabolic, but it is approximated by an isosceles triangle with small equal angles.

The extend length is 3 km + 12 cm, which is
3000 + 0.12 = 3000.12 m

Half of the extended length is 1500.06 m

Let h+ the rise of the center of the rail above ground.
From the Pythagorean theorem, obtain
h² + 1500² = 1500.6²
h² = 1500.06² - 1500² = 180.004
h = 13.4 m, which is about 0.9% of the original length.

Answer: 13.4 m