Based on the given equipotential lines, what is the direction of the electric field at point 2? b g c h e a d f

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Chapter1: Units, Trigonometry. And Vectors
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Based on the given equipotential lines, what is the direction of the electric field at point 2?

b

g

c

h

e

a

d

f

The image is a graph depicting equipotential lines with points labeled 1, 2, and 3. The graph is framed by a grid with units marked in volts (V) on the horizontal and vertical boundaries. 

- The horizontal axis ranges from 0 V to 100 V.
- The vertical axis also ranges from 0 V to 100 V.

There are three dashed, green equipotential lines shown on the graph:

1. The first equipotential line starts near the bottom-left corner and curves gently upwards to the right.
2. The second equipotential line is positioned above the first, bending more sharply upwards as it moves from left to right.
3. The third equipotential line is at the topmost position and curves the steepest upwards.

Key Points:
- Point 1 is located at 50 V on the vertical axis.
- Point 2 is positioned at 50 V on the horizontal axis.
- Point 3 is situated near the middle of the graph, between Points 1 and 2, slightly above the center.

Each grid square represents a 1 cm by 1 cm section, as indicated by the scale at the top-left corner of the graph.
Transcribed Image Text:The image is a graph depicting equipotential lines with points labeled 1, 2, and 3. The graph is framed by a grid with units marked in volts (V) on the horizontal and vertical boundaries. - The horizontal axis ranges from 0 V to 100 V. - The vertical axis also ranges from 0 V to 100 V. There are three dashed, green equipotential lines shown on the graph: 1. The first equipotential line starts near the bottom-left corner and curves gently upwards to the right. 2. The second equipotential line is positioned above the first, bending more sharply upwards as it moves from left to right. 3. The third equipotential line is at the topmost position and curves the steepest upwards. Key Points: - Point 1 is located at 50 V on the vertical axis. - Point 2 is positioned at 50 V on the horizontal axis. - Point 3 is situated near the middle of the graph, between Points 1 and 2, slightly above the center. Each grid square represents a 1 cm by 1 cm section, as indicated by the scale at the top-left corner of the graph.
The image shows a circular diagram with eight arrows radiating outward from a central point, resembling a compass rose. Each arrow is labeled with a lowercase letter (a through h), representing different directions:

- Arrow "a" points upwards.
- Arrow "b" points to the upper right.
- Arrow "c" points directly to the right.
- Arrow "d" points to the lower right.
- Arrow "e" points downwards.
- Arrow "f" points to the lower left.
- Arrow "g" points directly to the left.
- Arrow "h" points to the upper left.

This diagram can be used to illustrate directions or vectors in mathematical or physics contexts, helping students understand concepts like directionality and vector components.
Transcribed Image Text:The image shows a circular diagram with eight arrows radiating outward from a central point, resembling a compass rose. Each arrow is labeled with a lowercase letter (a through h), representing different directions: - Arrow "a" points upwards. - Arrow "b" points to the upper right. - Arrow "c" points directly to the right. - Arrow "d" points to the lower right. - Arrow "e" points downwards. - Arrow "f" points to the lower left. - Arrow "g" points directly to the left. - Arrow "h" points to the upper left. This diagram can be used to illustrate directions or vectors in mathematical or physics contexts, helping students understand concepts like directionality and vector components.
Expert Solution
Step 1 ans. g

Electric field /electric field line s are always perpendicular to the equipotential surface s.

At point 2 the  equipotential surface has a potential 50 volts and increasing potential is towards   +ve x axis and decreasing potential is towards -ve X axis. So direction of electric field is towards decreasing potential

So direction of electric field / electric field line is   parallel towards -ve X axis.

And magnitude of electric field can be calculated as _ve gradiant of potential function.

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