Estimate the electric field at the locations a and bon the 10 v equipotential in

Question

Estimate the electric field at the locations a and bon the 10 v equipotential in your equipo map, shown in Figure 2. Again create nearby are approximately t 1 away for (approximately 9v and 11 V-Estimate that values) and Ar between your by calculating Av (using the provided equipotential perpendicular distance along new equipotentials, using the labels for potential and measure for a line can, and try to see how small through each again, try to be as accurate as you field independently differences in affect the resulting the your at the 10V and 15 v locations using an E-field sensor. Record and compare Magnitude of E-field estimated from El.location a Magnitude of E-field estimated using sensor, location at Confirm the estimated and measured values are in rough agreement. Magnitude of E-field estimated from location b: Magnitude of E-field estimated using sensor, location bc Confirm the estimated and measured values are in rough agreement. Did you prediction agree with your measurement? Briefly describe: Using Equation 3, briefly explain your observation. Which field magnitude is larger, and why?

Explanation / Answer

Solution :-

Given

Voltage = 10 V

We have to create equipotentials that are approximately+- 1V away for (i.e limit varies from 9 V to 11 V)

a) V/ r is actually correlated with magnitude of E field in the following manner:-

E field = -(V/ x i + V/ y j + V/ z k)

Considering one dimensional field and r between new equipotentials, the relation will be as follows:-

E field = - V/ r

Here V is the change in the potential which occurs over a change in location r

Now in terms of limit we have to integrate both sides, then

(From 9 to 11) E field = - V/ r

(E^2/2) ((From 9 to 11) = -V/r

So

11^2/2- 9^2/2 = -V/r

So magnitude of electric field = 60.5 - 40.5 = 20.5 V/m

b) Magnitude of field using sensor location is almost as close as calcuated value varies from (+-) 1

Hnece the Magnitude of field using sensor location = 19.5 to 21.5 V/m

c) Since the calculated values

i.e. 20.5 - 19.5 = 1 V/m and

21.5 - 20.5 = 1V/m

Hence measure value and calculated value are in rough agreement.

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.