Answer:

Explanation:
As we know that electric field due to infinite line charge distribution at some distance from it is given as

now we need to find the electric field at mid point of two wires
So here we need to add the field due to two wires as they are oppositely charged
Now we will have

now plug in all data



now we have



Answer:
68.9233231661
Explanation:
Just put it into your calculator, shift sin should do it but it will come up like this:
which is the same as arcsin
Answer:
Moving a unit "positive" test charge from A to B will result in a reduction in potential
V = K Q / R potential at a point
V2 - V1 = K Q (1 / .4 - 1 / .15) = = k Q (.15 - .4) / .06 = -4.17 K Q
V2 - V1 = -4.17 * 9 & 10E9 * 6.25 E-8
V2 - V1 = -4.17 * 562.5 J/C
V = - 2346 Volts
Answer:
Send a pic of it so i can mark it.
Explanation:
Answer:
I believe the answer is Bromine forms covalent bonds because it has many electron shells, but neon has only two electron shells and is tightly bound to its electrons.
Explanation:
valence electrons are the outermost shell, so when you go through keeping that in mind it helps you find the right answer