Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.?
Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.
(a) What is the electric potential at point a due to q1 and q2?
(b) What is the electric potential at point b?
(c) A point charge q3 = -6.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?
Answer:
a) the potential is zero at the center .
Explanation:
a) since the two equal-magnitude and oppositely charged particles are equidistant
b)(b) Electric potential at point b, v = Σ kQ/r
r = 5cm = 0.05m
k = 8.99*10^9 N·m²/C²
Q = -2 microcoulomb
v= (8.99*10^9) * (2*10^-6) * (1/√2m - 1) / 0.0500m
v = -105 324 V
c)workdone = charge * potential
work = -6.00µC * -105324V
work = 0.632 J
Energy absorbed by the radiologist is 1.775*10^-6J.
To find the answer, we have to know about the radiation.
<h3>How much energy has she absorbed?</h3>
- We have the expression for dose of absorption as,
where; E is the energy absorbed and m is the mass of the body.
- From the above expression, substituting appropriate values given in the question, we get,
Thus, we can conclude that, the Energy absorbed by the radiologist is 1.775*10^-6J.
<h3 />
Learn more about the radiation here:
brainly.com/question/24491547
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Answer:
i think it would be B. Aluminum, Al and D. Boron, B
Explanation:
since they're both in group 13 and they forms a 3+ ion
The vesicles release neurotransmitters. These cross the synapse and are accepted by the receptors in the dendrites of the next neuron.
Explanation:
An axon, or nerve fiber, is a long slender projection of a nerve cell, or neuron, that conducts electrical impulses away from the neuron's cell body. Axons are in effect the primary transmission lines of the nervous system, and as bundles they help make up nerves.
When an action potential reaches the axon terminal, it depolarizes the membrane and opens voltage-gated Na+ channels. Na+ ions enter the cell, further depolarizing the presynaptic membrane.
