The quantity of charge q (in coulombs) that has passed through a surface of area 2.08 cm2 varies with time according to the equation q = 4t3 + 5t + 6, where t is in seconds.
(a) What is the instantaneous current through the surface at t = 1.08 s?
(b) What is the value of the current density?
To find the instantaneous current through the surface at t = 1.08 s, we need to find the derivative of the charge with respect to time and evaluate it at t = 1.08 s. Let's start by finding the derivative.
q = 4t^3 + 5t + 6
To find the derivative of q with respect to t, we differentiate each term separately:
dq/dt = d/dt(4t^3) + d/dt(5t) + d/dt(6)
The derivative of 4t^3 with respect to t can be found using the power rule for differentiation. For a term of the form ax^n, where a is a constant and n is a power, the derivative is given by nax^(n-1).
Taking the derivatives of each term, we get:
dq/dt = 12t^2 + 5
Now, let's evaluate the expression dq/dt at t = 1.08 s:
dq/dt = 12(1.08)^2 + 5
= 14.796
Therefore, the instantaneous current through the surface at t = 1.08 s is approximately 14.796 A (amperes).
Now, let's move on to finding the current density.
Current density (J) is defined as the current per unit area. In this case, we need to divide the instantaneous current (q) by the surface area (A).
We are given that the surface area is 2.08 cm^2. However, it is more convenient to work with SI units, so let's convert the surface area to square meters.
1 cm^2 = (1/100)^2 m^2 = 0.0001 m^2
So, the surface area in square meters is:
A = 2.08 cm^2 * 0.0001 m^2/cm^2
= 0.000208 m^2
Now, we can calculate the current density:
J = q / A
= (4t^3 + 5t + 6) / 0.000208
To find the value of the current density, we need to know the value of t. If you provide a specific value of t, I can plug it into the equation and calculate the current density for you.