Class Management I Help Practice Problems 4 Potential Begin Date: 2/8/2016 12:00

Question

Class Management I Help Practice Problems 4 Potential Begin Date: 2/8/2016 12:00:00 AM Due Date: 5/7/2016 12:00:00 AM End Date: 5/7/2016 12:00:00 AM (10%) Problem 5: A rod of length L-0.55 m is placed along the x-axis with its y-axis center at the origin. The rod has a non-uniform linear charge density of kx C/m where k 6. E-12 C/m is a constant factor Job X-axis otheexpertta.com 100% Part (a) What is the value of the potential at the point (0.yob) in volts if yob 3.05 m? Grade Summary Deductions Potential 84 8 9 HOME Submissions asin() Attempts remaining 4 5 6 acos per attempt) atan() acotano sinh() detailed view cosh00 tanho cotanho END Degrees Radians NO BACKSPACE DEL CLEAR submit at Feedback I give up! Hints: 4 for a deduction. Hints remaining: 0 Feedback 5% deduction per feedback. -Look for variables that are constant. They simplify calculus How do da, 9, and the coordinates relate? Use the general form of the potential with a charge distribution. -You may have to split the integration into two parts to deal with the absolute value

Explanation / Answer

Lets calculate the potential due to right half, i.e. x varies from 0 to 0.55/2=0.275m. Take a small element at x=x and dx length.

dq = lambda.dx =6.9E-12 *x*dx

r = sqare root of (x2 + yobs2) = (x2 +3.052)0.5 = (x2 +9.3)0.5 =

Potential = integral of Kdq/r

= 2*integral of [9E9 * 6.9E-12 *x*dx]/(x2 +9.3)0.5 =

=0.1242* integral of xdx/(x2 +9.3)0.5

Let x2 +9.3=z, dz=2xdx

potential =0.5*0.1242 integral of dz/z0.5=

= 0.1242* z0.5=0.1242 *(x2 +9.3)0.5 =

taking value from x=0 to x=.275

potential =0.12425(3.062372- 3.05)= 0.1242 * 0.012372 = 0.001536 V answer

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