Answer:
Explanation:
We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .
∫ E ds = q / ε
Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so
∫ E ds represents total electric flux passing through the closed surface due to charge q enclosed in the surface .
This also represents total flux coming out of the charge q on all sides .
This is equal to q / ε where ε is a constant called permittivity which depends upon the medium enclosing the charge . For air , its value is 8.85 x 10⁻¹² .
If charge remains the same but radius of the sphere enclosing the charge is doubled , the flux coming out of charge will remain the same .
It is so because flux coming out of charge q is q / ε . It does not depend upon surface area enclosing the charge . It depends upon two factors
1 ) charge q and
2 ) the permittivity of medium ε around .
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Answer:
B) Gets smaller
Explanation:
The difference of phase between current and voltage in a AC circuit is the phase angle and it depends on the value of Z ( circuit impedance)
Z = R + X where R is the resistive component and X the reactance component, which is due either to a presence of an inductor or a capacitor. In any case the total impedance depends on R the resistive, and the phase angle φ is:
tan⁻¹ φ = X/R
Have a look to a pure capactive circuit (we are talking about AC current) in this case current leads voltage by 90⁰. If we add a resistor in the circuit the current still will lead a voltage but in this condition the phase angle will be smaller,
If R increase, X/R decrease and tan⁻¹ φ also decrease
Answer:
B-flood
Explanation:
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