The electric flux is phi = alpha sigma + beta | vedclass

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(C) Given the equation for electric flux: $\phi = \alpha \sigma + \beta \lambda$.By the principle of homogeneity of dimensions,the dimensions of each

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displacement

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(C) Given the equation for electric flux: $\phi = \alpha \sigma + \beta \lambda$.By the principle of homogeneity of dimensions,the dimensions of each term must be equal: $[\phi] = [\alpha \sigma] = [\beta \lambda]$.From $[\phi] = [\alpha \sigma]$,we get $[\alpha] = \frac{[\phi]}{[\sigma]}$.From $[\phi] = [\beta \lambda]$,we get $[\beta] = \frac{[\phi]}{[\lambda]}$.Now,consider the ratio $\frac{\alpha}{\beta}$: $\left[\frac{\alpha}{\beta}\right] = \frac{[\phi]/[\sigma]}{[\phi]/[\lambda]} = \frac{[\lambda]}{[\sigma]}$.The dimensions of linear charge density $\lambda$ are $[Q/L]$ and surface charge density $\sigma$ are $[Q/L^2]$.Substituting these: $\left[\frac{\alpha}{\beta}\right] = \frac{[Q/L]}{[Q/L^2]} = \frac{L^2}{L} = [L]$.Since the dimension is $[L]$,the ratio $\left(\frac{\alpha}{\beta}\right)$ represents length,which is a unit of displacement.

The electric flux is $\phi = \alpha \sigma + \beta \lambda$,where $\lambda$ and $\sigma$ are linear and surface charge density,respectively. The ratio $\left(\frac{\alpha}{\beta}\right)$ represents:

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