A 10-cm-long thin glass rod uniformly charged to 15.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 15.0 nCare placed side by side, 4.50 cm apart. What are the electric field strengths E1 to E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?

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Asked Sep 6, 2019
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A 10-cm-long thin glass rod uniformly charged to 15.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 15.0 nCare placed side by side, 4.50 cm apart. What are the electric field strengths E1 to E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?

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Expert Answer

Step 1

Given:

Length of each rod = 10 cm

Charge in glass rod = 15 nC

Charge on pastic rod = - 15 nC

Distance between the rods = 4.5 cm

Step 2

Writing the Electric field at a distance ‘r’ in front of the charged rod:

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1/2 Enet If the length of the rod is L and charge on the rod is +Q, then Es (sin e)) 2 7TE Where Q is the charge on the rod, r' is the distance from the rod and is the angle as shown in the figure. (The direction of net electric is away from the rod in case of positive charge and towards the rod in case of negative charge on the rod) Note: sin e (2) +(r)

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Step 3

Calculating the electric field between the rods at a distance 1 cm from the glass rod:

Let&rsquo...

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Qrias(sin .(3) Electric field due to glass rod, Era 2 Opizte(sine) --.(4) 2E Electric field due to pastic rod, Epiaze pastec (5x10 m) NSx10m)*+(]x10-m) sin epiers - 0.98 , sin e = 0.98 (5x10m) (5x10 m +(3.5x10 m) 0.82, sineie 0.82 pastic astic Substitue Qlass15 nC 15x 109 C, rss1 cm 1x 102m and sin ia =0.98 in eq 3 15x10 Eot 2(3.14)(8.85x1012x1x10-2 (0.98) E 26.5x10 N/c 15x 10-9 C, Tplastic 3.5 cm 3.5x 102m and sin e Substitue Qplastic -15 nC 0.82 pastie -15 x10 (0.82) Flazt 23.14)8.85 x103.5x10) Eo6.3x10 N/C Net electric field, E = Ei +E (26.5x10-6.3x10)Nc pztie E 2x10 N/C (Towards +ve x-axis)

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Electric Charges and Fields

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