A student of 4 feet tall went for swimming in a pool. He saw the depth of water in the pole less than 4 feet .Will he sink , wri
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:
About 2.19 seconds
Explanation:
Since the ball is dropped from rest, there is no initial velocity, and you can write the following equation:
Hope this helps!
Question is missing. Found on internet the complete text of the problem:
"<span>Two very large parallel sheets are 5.00 cm apart. Sheet A carries a uniform surface charge density of −9.70µC/m2, and sheet B, which is to the right or A, carries a uniform charge density of −11.5 µC/m2. Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 cm to the right of sheet A; (b) 4.00 cm to the left of sheet A; (c) 4.00 cm to the right of sheet B."
Solution:
(a) The electric field produced by a uniformly charged sheet at any distance is given by
</span>
where
is the charge density and 
is the vacuum permittivity.
First of all, let's compute the fields generated by the two sheets separately. The two densities of charge are
and 
.
Sheet a gives an electric field of
where the negative sign means the field points towards sheet A, in any point of the space.
The electric field produced by sheet B is given by:
and again, the negative sign means that the field at any point of the space points towards sheet B.
The point at which we have to compute the total field is at 4.00 cm right of sheet A. Since the two sheets are 5.00cm far apart, it means that this point is between the two sheets. Therefore, in this point the two fields point into opposite directions. Therefore, the total field is
And the direction is towards sheet B, since it has a field with stronger intensity.
(b) Field at 4.00 cm to the left of sheet A: in this point of the space, the two fields point towards same direction (on the right, towards both sheet A and sheet B). So, the total field is simply the sum of the two fields:
towards right.
(c) Field at 4.00 cm to the right of sheet B. As before, the two fields in this point have same direction (both towards left, pointing towards both sheet A and sheet B). And so, the total field is simply the sum of the two fields:
towards left.
"<span>Two very large parallel sheets are 5.00 cm apart. Sheet A carries a uniform surface charge density of −9.70µC/m2, and sheet B, which is to the right or A, carries a uniform charge density of −11.5 µC/m2. Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 cm to the right of sheet A; (b) 4.00 cm to the left of sheet A; (c) 4.00 cm to the right of sheet B."
Solution:
(a) The electric field produced by a uniformly charged sheet at any distance is given by
</span>
where
First of all, let's compute the fields generated by the two sheets separately. The two densities of charge are
Sheet a gives an electric field of
where the negative sign means the field points towards sheet A, in any point of the space.
The electric field produced by sheet B is given by:
and again, the negative sign means that the field at any point of the space points towards sheet B.
The point at which we have to compute the total field is at 4.00 cm right of sheet A. Since the two sheets are 5.00cm far apart, it means that this point is between the two sheets. Therefore, in this point the two fields point into opposite directions. Therefore, the total field is
And the direction is towards sheet B, since it has a field with stronger intensity.
(b) Field at 4.00 cm to the left of sheet A: in this point of the space, the two fields point towards same direction (on the right, towards both sheet A and sheet B). So, the total field is simply the sum of the two fields:
towards right.
(c) Field at 4.00 cm to the right of sheet B. As before, the two fields in this point have same direction (both towards left, pointing towards both sheet A and sheet B). And so, the total field is simply the sum of the two fields:
towards left.