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Electric fields describe the interaction of stationary charged matter. They underlie the working of diverse technology from atom smashers to the poor cell reception you're getting right now.

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A uniform electric field of \( 200 \text{ N/C} \) points to the left as shown in above figure. When the distance between points \( A \) and \( B\) is \( d= 5 \text{ cm}, \) what is the difference in potential \( V_A - V_B \) between points \( A \) and \( B? \)

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A charge of \( 6.0 \text{ nC}, \) is initially at a point that is \( r_1 = 3.0 \text{ m}, \) away from a charge of \( 1.0 \text{ nC} \) moves further away to a point where the distance is \(r_2= 7.0 \text{ m}. \) What is the approximate potential difference between the two points.

Assume that electric constant is \( \epsilon_0 = 8.9 \times 10^{-12} \text{ F/m}. \)

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A 9V battery has an electric potential difference of \(9~\mbox{V}\) between the positive and negative terminals. How much kinetic energy **in J** would an electron gain if it moved from the negative terminal to the positive one?

**Details and assumptions**

- The charge on the electron is \(-1.6 \times 10^{-19}~\mbox{C}\).
- You may assume energy is conserved (so no drag or energy loss due to resistance for the electron).

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As shown in the above figure, three point charges \( q_1 = 2.0 \text{ nC}, q_2 = 5.0 \text{ nC} \) and \( q_3 = 4.0 \text{ nC}\) are placed at the three corners of a square with side length \( d = 7 \text{ m}. \) Find the approximate potential at the point \( A. \)

Assume that electric constant is \( \epsilon_0 = 8.9 \times 10^{-12} \text{ F/m} \) and \( \sqrt{2} \) is \( 1.4. \)

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