Finding the electric field from equipo 1 cm× 1 cm grid is superimposed on an EXA

Question

Finding the electric field from equipo 1 cm× 1 cm grid is superimposed on an EXAMPLE 21.9 In FIGURE 21.23 a 1 equipotential map of the potential. Estimate the strength and direction of the electric field at points 1, 2, and 3. Show you results graphically by drawing the electric equipotential map. FIGURE 21.23 Equipotential lines. field vectors on th 1 cm 0V 50 V 100 V 1 cm 100 V- PREPARE The electric field is perpendicular to the equipoten es, points "downhill," and depends on the spacing between equipotential lines. The potential is highest on the bottom right. An elevation graph of the potential would look like the er-right quarter of a bowl or a football stadium. IStant but unseen source charges have create

Explanation / Answer

We know that, electric field is related to electric potential as:

E = V/d

for the point 1 in the figure,

V = 50 Volts, d = 1 cm

E = 50 V/1 cm = 50 V/cm = 5000 V/m

Hence, E = 50 V/cm

2)The electric field points towards higher potential to lower potential.

3)For every 2 cm, there is a rise of 1 V in the graph,

E = 1 V/2 cm = 0.50 V/cm

Hence, E = 0.50 V/cm

4)v = 1.905 V ; d = 1 in = 2.54 cm

E = V/d = 1.905 V /2.54 cm = 0.75 V/cm

Hence, E = 0.75 V/cm

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