When you reverse the direction of the current, the current loop generated by the magentic field is revered.
Answer:
90 C
Explanation:
Electric current: This can be defined as the rate of flow of electric charge in a circuit. This can be expressed mathematically as,
I = dQ/dt
dQ = Idt
∫dQ = ∫Idt
Q = It................................ Equation 1
Where Q = amount of charge, I = current, t = time.
Given: I = 3.6 A, t = 25 s.
Substituting into equation 1,
Q = 3.6(25)
Q = 90 C.
Hence the amount of charge passing through the cross section of the conductor = 90 C
Answer:
C = 4,174 10³ V / m^{3/4}
, E = 7.19 10² / ∛x, E = 1.5 10³ N/C
Explanation:
For this exercise we can calculate the value of the constant and the electric field produced,
Let's start by calculating the value of the constant C
V = C
C = V / x^{4/3}
C = 220 / (11 10⁻²)^{4/3}
C = 4,174 10³ V / m^{3/4}
To calculate the electric field we use the expression
V = E dx
E = dx / V
E = ∫ dx / C x^{4/3}
E = 1 / C x^{-1/3} / (- 1/3)
E = 1 / C (-3 / x^{1/3})
We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E
E = 3 / C (0- (-1 / x^{1/3}))
E = 3 / 4,174 10³ (1 / x^{1/3})
E = 7.19 10² / ∛x
for x = 0.110 cm
E = 7.19 10² /∛0.11
E = 1.5 10³ N/C
EC_1 + EP_1 = EC2 + EP_2
EC_2 = 0
EC_2 = EP_1 - EP_2
EC_2 = mg(H_1 - H_2) = 0.20 kg * 9.8 m/s^2 * (3.25 m - 1.5m) = 3.43 J
Answer:

Explanation:
We are given that
Surface area of membrane=
Thickness of membrane=
Assume that membrane behave like a parallel plate capacitor.
Dielectric constant=5.9
Potential difference between surfaces=85.9 mV
We have to find the charge resides on the outer surface of membrane.
Capacitance between parallel plate capacitor is given by

Substitute the values then we get
Capacitance between parallel plate capacitor=

V=


Hence, the charge resides on the outer surface=