**Electric Fields and Potentials: Problem Explanation**There is a positive charge \( Q \) at \( x = -5 \) cm on the x-axis, and a negative charge \(-2Q\) at \( x = 15 \) cm on the x-axis. What are the total electric field and electric potential at \( x = 10 \) cm on the x-axis? Assume the usual convention for defining the potential of a point charge.### Options for Answer:1. \(\vec{E} \) points to positive x, \(V\) is negative2. \(\vec{E} \) points to positive x, \( V\) is positive3. \(\vec{E} \) points to negative x, \( V\) is negative4. \(\vec{E} \) is zero, \( V\) is zero5. \(\vec{E} \) is zero, \( V\) is negative6. \(\vec{E} \) points to negative x, \( V\) is positive**Correct Answer:**3. \(\vec{E} \) points to negative x, \( V\) is negative### Explanation:This problem asks us to determine both the direction of the electric field vector (\(\vec{E}\)) and the sign of the electric potential (\(V\)) at a specific point on the x-axis due to the presence of two point charges. Based on electrostatics, the direction and magnitude of the electric field and the potential can be calculated using the positions and values of these charges.

Given: The charge Q at a distance r1 = 5 x 10-2 m The charge -2Q at a distance r2 = 15 x 10-2 cm

Answered: There is a positive charge Q at x = 5…

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