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Figure shows two rods each of length l placed side by side, with their facing ends separateed by a distance a.Chrges +q, -q reside on the rods as shown. Calculate the electric force on the rod A due to the rod B. Discuss the cases when lgtgta, agtgt1 |
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Answer» Solution :The FORCE on the rod A due to the charge +q of the rod B =- q^2/(4PI eplsilon_0(l+a0^2)+q^2/(4 pi eplsilon_0 a^2)` TOWARDS right. The force on this rod due to the charge -q. `q^2/(4 pi eplsilon_0(2l+a)^2) - q^2/(4 pi epsilon_0(l+a)^2)` towards right. The resultan force on the rod is `F= q^2 /(4 pi epsilon_0) [1/a^2-2/((l+a)^2)+1/((2l+a)^2)`]` towards right. If `lgtgta`, the last two terms in the square bracket are negligible as COMPARED to the first term. then, `F~~q^2/(4pi epsilon_0 a^2)` If `agtgt1` `F=q^2/(4 pi eplsilon_0) [1/a^2-2/a^2+1/a^2]~~0` Two neutral OBJECT placed far away exert only negligible force on each other but when they are placed closer they may exert appreciable force. |
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