Answer:
The magnitude of the flux of electric field through a square of surface area is zero.
Explanation:
It is given that square box is parallel to yz-plane which has normal vector perpendicular to plane in x-direction. Angle between normal vector of area and electric field is 90°. Substituting in (1)
The electric field at one corner of a square is 1614217 N/C.
Explanation:
The distance between x and y direction diagonals.
As per the given details the distance between diagonals is calculated as
0.5² + 0.5² = c² => c = 0.707 m
Charge to the right: In x direction
In order to find the electric charge towards x direction
we use e = kq/r² formula
As 'k' is coulomb's constant it's value is 9 x N m²/C²
e = (9 x )(250 x
) / (0.5)²
e = 9 x N/C
Charge diagonal:
e = kq/r²
e = [(9 x )(250 x
) / (0.707)²] cos 45
e = 225000√2 N/C
X direction sum = 1218198 N/C.
Similarly as shown in x direction the charge is same for y direction also
Charge below: For y direction
e = kq/r²
e = (9 x )(250 x
) / (0.5)²
e = 9 x N/C
Charge diagonal:
e = kq/r²
e = [(9 x )(250 x
) / (0.5)²] sin 45
e = 159099 N/C
Y direction sum = 1059099 N/C
Resultant electric field strength:
1218198 ² + 1059099² = e²
e = 1614217 N/C [45 degrees below the horizontal]
1) Current in each bulb: 0.1 A
The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:
And so, the current through the circuit is (using Ohm's law):
And since the two bulbs are connected in series, the current through each bulb is the same.
2) 4 W and 8 W
The power dissipated by each bulb is given by the formula:
where I is the current and R is the resistance.
For the first bulb:
For the second bulb:
3) 12 W
The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so: