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Sveta_85 [38]
3 years ago
12

What potential difference is needed to stop an electron that has an initial velocity v=6.0

Physics
1 answer:
Assoli18 [71]3 years ago
7 0

Formula for kinetic energy of an object:

KE = 0.5mv²

m is the mass and v is the velocity.

Formula for the work done on a charged object by moving it through a potential difference:

W = ΔVq

ΔV is the potential difference and q is the charge of the object.

To find the potential difference needed to decelerate an electron to rest, set the work done on the electron equal to its kinetic energy:

W = KE

Substitute W = ΔVq and KE = 0.5mv²

ΔVq = 0.5mv²

Given values:

q = 1.6×10⁻¹⁹C

m = 9.11×10⁻³¹kg

v = 6.0m/s

Plug in the given values and solve for ΔV

ΔV×1.6×10⁻¹⁹ = 0.5×9.11×10⁻³¹×6.0²

ΔV = 1.02×10⁻¹⁰V

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\blue{\bold{\underline{\underline{Answer:}}}}

  • \green{\tt{Work\:done=24\:J}}

\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}

\green{\underline{\bold{Given :}}} \\  \tt: \implies Constant \: force(F) = 8 \: N \\  \\ \tt: \implies Displacement(s) = 3 \: m \\  \\ \red{\underline{\bold{To \: Find : }}} \\  \tt:  \implies Work \: Done(W.D) = ?

• <u>According to given question</u> :

\green{ \star} \tt \:  \theta \:  = 0 \degree \:  \:  \:  \: (Angle \: between \: force \: and \: displacement) \\  \\  \bold{As \: we \: know \: that} \\  \tt:  \implies Work \: Done = FS \: cos  \: \theta \\  \\  \tt:  \implies Work \: Done = 8 \times 3 \times cos \:0 \degree \\  \\ \green{ \circ} \tt \: cos  \: 0 \degree = 1  \\  \\  \tt:  \implies Work \: Done =24 \times 1 \\  \\   \green{\tt:  \implies Work \: Done =24 \: J}

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