Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k 
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]
Answer:
I = 24 A
Explanation:
This is Parallel Circuit and it is the first principle of parallel circuit that voltage will be equal in all components in the circuit
It includes 10 resistors Therefore the voltage across,
R1 = R2 = R3 = R4 = R5 = R6 = R7 = R8 = R9 = R10 = voltage in battery
<h2>
Ohm's Law</h2>
We will apply Ohm's Law to each resistor to find its current because we know the voltage across each resistor is 12 V and the resistance of each resistor is 5Ω
I (R1) = E (R1) / R1
I (R1) = 12v / 5Ω
I (R1) = 2.4 A
The value resistance E of all resistors are same therefore by applying the formula above the value of current in all resistors will be 2.4 A
The Total current in the circuit will be
I (total) = I (1) + I (2) + I (3) + I (4) + I (5) + I (6) + I (7) + I (8) + I (9) + I (10)
I (total) = 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4
I (total) = 24 A
Answer:-q
Explanation:
Given
Capacitor is charged to a battery and capacitor acquired a charge of q i.e.
+q on Positive Plate and -q on negative Plate.
If the plate area is doubled and the plate separation is reduced to half its initial separation then capacitor becomes four times of initial value because capacitor is given by
where A=area of capacitor plate
d=Separation between plates
This change in capacitance changes the Potential such that new charge on the negative plate will remain same -q
The only information you would need to decide if the can will float is the density of the can, which requires knowing the mass and volume. If the density of the can is less than one, the can will float. if it is greater than one, it will not float, as water's density is one.
Answer:
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Explanation:
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