Electric Field and Potential

Physics Experiment by Anshuman Swain, SR No. 10758

Experiment: Electric field and potential

By Anshuman Swain, SR No. 10758, Group-C

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A: Map the electric potential and field around a charged sphere. Refer to Leybold manual P3.1.3.4.

Experiment: Choose an appropriate value of the High Voltage source (e.g. 4kV). Charge the sphere by connecting it to the source. Measure the electric potential using the flame probe at four places at equal distance r from the centre of the sphere. Verify that the potential is uniform with r.

Voltage of HV source = 5 kV

r=18.6 cm

V0(in kV) Vπ/2(in kV) Vπ(in kV) V3π/2(in kV)4.3 4.4 4.4 4.4

E0(in kV/m) Eπ/2(in kV/m) Eπ(in kV/m) E3π/2(in kV/m)64.3 64.4 64.6 64.7

Conclusion: Here we see that the potential is uniform with r. We also see that electric field is also constant with distance kept

Measure the potential as a function of distance from the centre of the charged sphere. Plot it versus 1/r to confirm that it is a straight line.

r(cm) V(kV) 1/r(cm-1) r’=(r1+r2)/2(cm) E(kV/m) 1/r’2(cm-2)5.0 16.50 0.200 5.25 330 0.0365.5 16.12 0.182 5.75 293 0.0306.0 15.60 0.167 6.25 260 0.0266.5 14.95 0.154 6.75 230 0.0226.5 13.33 0.154 6.75 205 0.0227.0 13.37 0.143 7.25 191 0.0197.5 13.88 0.133 7.75 185 0.0178.0 14.24 0.125 8.25 178 0.0158.5 13.69 0.118 8.75 161 0.0139.0 13.86 0.111 9.25 154 0.0129.5 14.35 0.105 9.75 151 0.01110.0 14.70 0.100 10.25 147 0.010

From the above data, find the electric field (this will be the field at the position r’ that is the average of the two distances r1 and r2 used to find the field) and plot is as a function of 1/r’2. Verify that it is a straight line.

Observation: Now from the data that has been recorded, we can plot the graph and see that the graph is nearly a straight line (within acceptable error limits). Here the error bars show the error in the measurement of the various lengths.

1 | P a g eElectric field and potential


130 180 230 280 330 3800

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

f(x) = 0.000134638946361337 x − 0.00846481514232688R² = 0.977978055378067

Electric Field (in kV/m)

1/r2

(in 1

/m2)

B: Map the electric potential and field inside a parallel plate capacitor. Refer to Leybold manual P3.1.3.3 and P3.1.7.4.

Choose an appropriate distance d between the two parallel plates (e.g. 5cm). d should be substantially smaller than the dimensions of the plates for the assumption of uniform field to hold. Keep one plate of the capacitor at a constant voltage (say 4kV) and connect the other plate to the grounded negative terminal of the high voltage source.

Study the potential parallel to the plates: Keeping the distance, a of the flame probe from the plates constant, measure the potential from one edge to the other.

a=7.5 cm

y(cm) 1.7 2.8 3.7 4.6 5.5 9.1 10.0 10.9V(kV) 2.01 2.02 2.11 2.16 2.28 2.24 2.20 2.18

Observation: Here we observe that the potential at a given x, goes on increasing first and then goes on decreasing and is the maximum in the middle.

Study the potential perpendicular to the plates and find the field strength from slope of V vs x. Measure the potential at different distances from one of the plates and plot the data to get the value of E and also to confirm that the field is constant between the two parallel plates.

x(cm) 1.5 2 3 4 5 6 7V(V) 335 730 1330 1720 2450 3120 3780

Observation: From the graph that was generated using the observed data, we can calculate the value of the E as the slope, which is 61.29 kV/m. And as also the value is nearly constant, we can conclude the electric field is nearly constant.



0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.080

500

1000

1500

2000

2500

3000

3500

4000

f(x) = 61237.9603399433 x − 569.688385269122R² = 0.995655687321834

Distance (in m)

Pote

ntial

Diff

eren

ce (i

n V)

Using the electric field meter find the field as a function of the applied voltage from the HV supply for a constant distance d between the plates.

Vapplied(kV) 5 4.5 4 3.5 3E(kV/m) 81 77.4 67.6 60.4 51.8

2.7 3.2 3.7 4.2 4.7 5.250

55

60

65

70

75

80

85

f(x) = 15.08 x + 7.31999999999999R² = 0.98517342723265

Voltage applied (in kV)

Elec

tric

field

(in

kV/m

)

The theoretical distance that we get from the above graph, obtained experimentally, is the inverse of the slope, which is 6.6 cm.

Here the actual experimental value of d=6.3 cm. The expected values of E are as follows:

Vapplied(kV) 5 4.5 4 3.5 3E(kV/m) 79.4 71.4 63.4 55.6 47.6



% difference 2.02 8.40 6.62 8.63 8.82

Measure the field strength as a function of d and plot E vs 1/d to see expected dependence.

d(cm) 5 6 7 8 9E(kV/m) 86 65.5 60.5 53.8 48.2

0.1 0.12 0.14 0.16 0.18 0.2 0.2244

49

54

59

64

69

74

79

84

89

f(x) = 405.899063237794 x + 2.26949683937968R² = 0.967554818232947

1/r (in cm-1)

Elec

tric

Fiel

d (in

kV/

m)

Observation: As we can see here that there is a linear dependence between the two entities plotted, as expected.



Answers:

1. The fan that is rotating in the electric field is grounded. There is a gold plated foil which exposes an area as a sinusoidal function of time, and is present at the back of the rotating fan and So the charge that is not picked up by the fan goes to the foil. As there is a change in the area, there is a change in the charge taken in. Using the Gauss’ Law ,

E . A=Qϵ 0⇒E . d A

dt= iϵ 0

Thus by measuring the current we can measure the Electric field.

2. The flame probe ionises the air and as a result there is a flow of charges. And thus by the measurement of the flow of charges from the ionised air, we can measure the potential at a particular point.


Electric Field and Potential

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