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3 years ago

Conventionally, the field strength around a charged object is the direction of the force acting on a .

Physics

2 answers:

damaskus [11]3 years ago

8 0

Answer:

Field Strength:

Any charged body has the capacity to effect any test charge that comes inside its field or region. It can be also defined as,

"The total amount or magnitude of force,F or intensity felt by any unit test charge when it enters an electromagnetic or electric field."

For an electric field its unit will be, volt per meter or simply V/m.

Explanation:

A unit test charge inside an electric field:

When a unit test charge enters a given parameters or area set by the charged particle then it will surely experience a force,F equal to the magnitude of that charge body and it will be different as it continues to move closer or far inside the field.

As, we have, "E=F/q",(where "F" is the field strength and "q" is the unit test charge placed inside the field).

ExtremeBDS [4]3 years ago

7 0

Unit positive test charge

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Explanation:

Given that,

Capacitor = 30μC

Resistor = 49.0Ω

Voltage = 30.0 V

Frequency = 60.0 Hz

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Using formula of impedance

$Z=\sqrt{R^2+X_{c}^2}$ .....(I)

We need to calculate the value of $X_{c}$

Using formula of $X_{c}$

$X_{c}=\dfrac{1}{2\pi f c}$

$X_{c}=\dfrac{1}{2\times\pi\times60.0\times30\times10^{-6}}$

$X_{c}=88.42\ \Omega$

Put the value of $X_{c}$ into the formula of impedance

$Z=\sqrt{(49.0)^2+(88.42)^2}$

$Z=101.08\ \Omega$

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Using formula of rms current

$I_{rms}=\dfrac{V}{Z}$

$I_{rms}=\dfrac{30.0}{101.08}$

$I_{rms}=0.30\ A$

The rms current in the circuit is 0.30 A.

(b). We need to calculate the rms voltage drop across the resistor

Using formula of rms voltage

$V_{rms}=I_{rms}\times R$

Put the value into the formula

$V_{rms}=0.30\times49.0$

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The rms voltage drop across the resistor is 14.7 V

(c). We need to calculate the rms voltage drop across the capacitor

Using formula of rms voltage

$V_{rms}=I_{rms}\times X_{C}$

$V_{rms}=0.30\times88.42$

$V_{rms}=26.53\ V$

The rms voltage drop across the capacitor is 26.53 V.

Hence, This is the required solution.

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