COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
2nd Edition
ISBN: 9781319414597
Author: Freedman
Publisher: MAC HIGHER
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Chapter 16, Problem 85QAP
To determine

(a)

The magnitude and direction of the electric field just outside surface of nucleus.

Expert Solution
Check Mark

Answer to Problem 85QAP

The magnitude of the electric field just outside surface of nucleus is, E=1.769×1021N/C and it directed outwards from the nucleus.

Explanation of Solution

Given:

Diameter of atom, D=1.0×1010m

Radius of atom, R=0.5×1010m

Diameter of nucleus, d=9.2×1016m

Radius of nucleus, r=4.6×1016m

Charge on electron, q=1.6×1019C

Mass of electron, m=9.1×1031kg

Formula used:

The electric field is given by,
  E=kqr2

Where,
  E =Electric field
  q = Electric charges
  r =Distance from charge

Calculation:

The electric field is given by,
  E=kqr2E=9× 109×26×1.6× 10 194.6×4.6× 10 30E=1.769×1021N/C

The electric field is directed outwards from the nucleus because nucleus acts as point source.

To determine

(b)

The magnitude and direction of the electric field at the distance of outermost electron.

Expert Solution
Check Mark

Answer to Problem 85QAP

The magnitude and direction of the electric field at the distance of outermost electron is, E=1.497×1013N/C

Explanation of Solution

Given:

Diameter of atom, D=1.0×1010m

Radius of atom, R=0.5×1010m

Diameter of nucleus, d=9.2×1015m

Radius of nucleus, r=4.6×1015m

Charge on electron, q=1.6×1019C

Mass of electron, m=9.1×1031kg

Formula used:

The electric field is given by,
  E=kqr2

Where,
  E =Electric field
  q = Electric charges
  r =Distance from charge

Calculation:

The electric field is given by,
  E=kqR2E=9× 109×26×1.6× 10 190.5×0.5× 10 20E=1.497×1013N/C

The electric field is directed outwards from the nucleus.

To determine

(c)

The magnitude and direction of the acceleration of the outermost electron.

Expert Solution
Check Mark

Answer to Problem 85QAP

The magnitude and direction of the acceleration of outermost electron is, a=2.632×1024m/s2 and it is directed outwards from the nucleus.

Explanation of Solution

Given:

Diameter of atom, D=1.0×1010m

Radius of atom, R=0.5×1010m

Diameter of nucleus, d=9.2×1015m

Radius of nucleus, r=4.6×1015m

Charge on electron, q=1.6×1019C

Mass of electron, m=9.1×1031kg

Electric field, E=1.497×1013N/C

Formula used:

The force acting on the ball is,
F=qE

Where,
  F =Electrostatic force
  q = Electric charge
  E =Electric field

Calculation:

The force is given by,
  F=qE...(1)

But we know that,
F=ma...(2)

From equation (1) and (2)
qE=maa=qEma=1.6× 10 19×1.497× 10 139.10× 10 31a=2.632×1024m/s2

The acceleration is directed outwards from the nucleus.


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Chapter 16 Solutions

COLLEGE PHYSICS LL W/ 6 MONTH ACCESS

Ch. 16 - Prob. 11QAPCh. 16 - Prob. 12QAPCh. 16 - Prob. 13QAPCh. 16 - Prob. 14QAPCh. 16 - Prob. 15QAPCh. 16 - Prob. 16QAPCh. 16 - Prob. 17QAPCh. 16 - Prob. 18QAPCh. 16 - Prob. 19QAPCh. 16 - Prob. 20QAPCh. 16 - Prob. 21QAPCh. 16 - Prob. 22QAPCh. 16 - Prob. 23QAPCh. 16 - Prob. 24QAPCh. 16 - Prob. 25QAPCh. 16 - Prob. 26QAPCh. 16 - Prob. 27QAPCh. 16 - Prob. 28QAPCh. 16 - Prob. 29QAPCh. 16 - Prob. 30QAPCh. 16 - Prob. 31QAPCh. 16 - Prob. 32QAPCh. 16 - Prob. 33QAPCh. 16 - Prob. 34QAPCh. 16 - Prob. 35QAPCh. 16 - Prob. 36QAPCh. 16 - Prob. 37QAPCh. 16 - Prob. 38QAPCh. 16 - Prob. 39QAPCh. 16 - Prob. 40QAPCh. 16 - Prob. 41QAPCh. 16 - Prob. 42QAPCh. 16 - Prob. 43QAPCh. 16 - Prob. 44QAPCh. 16 - Prob. 45QAPCh. 16 - Prob. 46QAPCh. 16 - Prob. 47QAPCh. 16 - Prob. 48QAPCh. 16 - Prob. 49QAPCh. 16 - Prob. 50QAPCh. 16 - Prob. 51QAPCh. 16 - Prob. 52QAPCh. 16 - Prob. 53QAPCh. 16 - Prob. 54QAPCh. 16 - Prob. 55QAPCh. 16 - Prob. 56QAPCh. 16 - Prob. 57QAPCh. 16 - Prob. 58QAPCh. 16 - Prob. 59QAPCh. 16 - Prob. 60QAPCh. 16 - Prob. 61QAPCh. 16 - Prob. 62QAPCh. 16 - Prob. 63QAPCh. 16 - Prob. 64QAPCh. 16 - Prob. 65QAPCh. 16 - Prob. 66QAPCh. 16 - Prob. 67QAPCh. 16 - Prob. 68QAPCh. 16 - Prob. 69QAPCh. 16 - Prob. 70QAPCh. 16 - Prob. 71QAPCh. 16 - Prob. 72QAPCh. 16 - Prob. 73QAPCh. 16 - Prob. 74QAPCh. 16 - Prob. 75QAPCh. 16 - Prob. 76QAPCh. 16 - Prob. 77QAPCh. 16 - Prob. 78QAPCh. 16 - Prob. 79QAPCh. 16 - Prob. 80QAPCh. 16 - Prob. 81QAPCh. 16 - Prob. 82QAPCh. 16 - Prob. 83QAPCh. 16 - Prob. 84QAPCh. 16 - Prob. 85QAPCh. 16 - Prob. 86QAPCh. 16 - Prob. 87QAPCh. 16 - Prob. 88QAPCh. 16 - Prob. 89QAPCh. 16 - Prob. 90QAPCh. 16 - Prob. 91QAP
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