Question: 7. Now, remove the negative charge, and drag two positive charges, placing them 1 m apart, as shown below. Electric Field +10 Sensors Q00

7. Now, remove the negative charge, and drag two positive charges, placing

7. Now, remove the negative charge, and drag two positive charges, placing them 1 m apart, as shown below. Electric Field +10 Sensors Q00 Voltage Values Grld D sev Let's look at the resulting electric field due to both charges. Recall that the electric field is a vector, so the net electric field is the vector sum of the electric fields due to each of the two charges. Question-8: Where is the magnitude of the electric field roughly equal to zero (other than very far away from the charges)? Question-9: Consider a point 0.5 m above the midpoint of the two charges. As you can verify by removing one of the positive charges, the electric field due to only one of the positive charges is about 18 V/m. What is the magnitude of the total electric field due to both charges at this location?

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