6. -3 points osUniPhys1 24.5 WA.046.Tutorial. My Notes Ask Your In the diagram b

Question

6. -3 points osUniPhys1 24.5 WA.046.Tutorial. My Notes Ask Your In the diagram below, each unit on the horizontal axis is 8.00 cm and each unit on the vertical s4.00 cm. The equipotential lines in a region of uniform electric fieid are indicated by the blue lines. (Note that the diagram is not drawn to scale.) (a) What is the direction of the electric field? (b) Determine the magnitude of the electric field in this region. VIm (c) Determine the shortest distance for which the change in potential is 3 V

Explanation / Answer

(a)

It will be option 2 because we know that equipotential surfaces are perpendicular to the electric field lines.

And when we move in the direction of electric field the potential decreases.

(b)

It is a truth universally acknowledged that:

(Electric field) = -(gradient of potential)

So the magnitude of the electric field is given by

|E| = V/d

Where d is the distance between two adjacent lines, and V is the difference of potential between two adjacent lines. d is easily obtained by pythagoras' theorem and a consideration of similar triangles (look at the triangle formed at the origin by one line of equipotential and the two axes.) The result is d = 32/80. So

|E| = 6/(32/80)

= 1.677V/cm = 167.7V/m
(c)
The shortest distance for which the change in potential is 3V is d/2 or 17.88 mm

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