Answer:
Hence the answer is E inside
.
Explanation:
E inside
so if r1 will be the same then
E
proportional to 1/R3
so if R become 2R
E becomes 1/8 of the initial electric field.
<u>Answer</u>
The combined displacement is 2km north
<u>Explanation</u>
Since displacement is a vector quantity, we take into account the direction.
Good for us all the displacement vectors are in the same dimension, so we can make north positive and south negative or vice-versa.
We now add to obtain,

This will simplify to

Therefore the combined displacement is 2km north
Answer:
A. Zero
Explanation:
Given data,
The charge of the test charge, q = 1 C
The distance the charge moved against the filed of intensity, x = 30 cm
= 0.3 m
The electric field intensity, E = 50 N/C
The energy stored in the charge at 0.3 m is given by the formula,
V = k q/r
Where,
= 9 x 10⁹ Nm²C⁻²
The charge is moved from the potential V₁ to V₂ at 30 cm
Substituting the given values in the above equation
V₁ = 9 x 10⁹ x 30 / 0.3
= 1.5 x 10¹² J
And,
V₂ = 1.5 x 10¹² J
The energy stored in it is,
W = V₂ - V₁
= 0
Hence, the energy stored in the charge is, W = 0
Answer:
q = C V charge on 1 capacitor
q = 1 * 10E-6 * 110 = 1.1 * 10E-4 C per capacitor
N = Q / q = 1 / 1.1 * 10E-4 = 9091 capacitors
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