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Artist 52 [7]
3 years ago
10

Would a positive test charge released fromrest move toward a region of higher or lower electric potential(compared to the electr

ic potential at the point where it isreleased)?The answer is lower electric potentialNow imagine that the sign of our testparticle is changed from positive to negative, but the electricpotential remains the same. Which of the following statements iscorrect?a. Thedirection of the force will change and it will point to regions ofhigher potential energy.b. Thedirection of the force will not change and it will point to regionsof higher potential energy.c. Thedirection of the force will not change and it will point to regionsof lower potential energy.d. Thedirection of the force will change and it will point to regions oflower potential energy.
Physics
1 answer:
ANEK [815]3 years ago
5 0

Answer:

b

Explanation:

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Find the electric field at a point midway between two charges of +40.0 x 10^-9 c and.+60.0 x 10^-9 c
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Missing part in the text: "...the charges are <span>separated by a distance of 30.0 cm."
</span>
Solution:
The point midway between the two charges is located 15.0 cm from one charge and 15.0 from the other charge. The electric field generated by each of the charges is
E=k_e \frac{q}{r^2}
where
ke is the Coulomb's constant
Q is the value of the charge
r is the distance of the point at which we calculate the field from the charge (so, in this problem, r=15.0 cm=0.15 m).

Let's calculate the electric field generated by the first charge:
E_1 = (8.99 \cdot 10^9 Nm^2 C^{-2} ) \frac{+40.0 \cdot 10^{-9} C}{(0.15 m)^2}=1.6 \cdot 10^4 N/C

While the electric field generated by the second charge is
E_2 = (8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{+60.0 \cdot 10^{-9} C}{(0.15 m)^2}=2.4 \cdot 10^4 N/C

Both charges are positive, this means that both electric fields are directed toward the charge. Therefore, at the point midway between the two charges the two electric fields have opposite direction, so the total electric field at that point is given by the difference between the two fields:
E=E_2 - E_1 = 2.4 \cdot 10^4 N/C - 1.6 \cdot 10^4 N/C = 8000 N/C
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Explanation:

hope this helps you dear friend.

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Answer:

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Answer:

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Thus, after being touched, the charge on the can is negative.

And, If not touched, the charge on the ballon remains positive.

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